Solving Quadratic Equations Using the Quadratic Formula

Download for free
Find the solutions in the app
Android iOS
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Communicate Your Answer
Exercise name Free?
Communicate Your Answer 3
Communicate Your Answer 4
Communicate Your Answer 5
Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
Monitoring Progress 10
Monitoring Progress 11
Monitoring Progress 12
Monitoring Progress 13
Monitoring Progress 14
Monitoring Progress 15
Monitoring Progress 16
Exercises
Exercise name Free?
Exercises 1 Any quadratic quadratic equation can be written in standard form. ax2+bx+c=0​​ In this form a, b, and c are real numbers, and a​=0. We can solve this general equation by completing the square. ax2+bx+c=0​↓Solve by completing the square​↓x=2a-b±b2−4ac​​​​ This gives a general result that can be used to solve any quadratic equation by just imputing the values of the parameters a, b, and c. Therefore, this result is known as the Quadratic Formula.
Exercises 2 Let's start by reviewing the form of the Quadratic Formula. x=2a-b±b2−4ac​​​ This formula can be used to solve any quadratic equation. Here, a, b, and c are real numbers, and a​=0. These represent the same parameters used in the standard form of a quadratic equation. ax2+bx+c=0.​ Note that the formula involves the radical term b2−4ac​. The sign of the radicand determines the number of real solutions that the quadratic equations has, and this is why this value is known as the discriminant of the Quadratic Formula.If b2−4ac>0, the equation has two real solutions. If b2−4ac=0, the equation has one real solution. If b2−4ac<0, the equation has no real solutions.
Exercises 3 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. x2=7xLHS−7x=RHS−7xx2−7x=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. x2−7x=0⇔1x2+(-7)x+0=0​ We see that a=1, b=-7, and c=0.
Exercises 4 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. x2−4x=-12LHS+12=RHS+12x2−4x+12=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. x2−4x+12=0⇔1x2+(-4)x+12=0​ We see that a=1, b=-4, and c=12.
Exercises 5 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. -2x2+1=5xLHS−5x=RHS−5x-2x2−5x+1=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. -2x2−5x+1=0⇔-2x2+(-5)x+1=0​ We see that a=-2, b=-5, and c=1.
Exercises 6 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 3x+2=4x2LHS−4x2=RHS−4x2-4x2+3x+2=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. -4x2+3x+2=0​ We see that a=-4, b=3, and c=2.
Exercises 7 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 4−3x=-x2+3xLHS+x2=RHS+x2x2+4−3x=3xLHS−3x=RHS−3xx2+4−6x=0Commutative Property of Additionx2−6x+4=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. x2−6x+4=0⇔1x2+(-6)x+4=0​ We see that a=1, b=-6, and c=4.
Exercises 8 We want to write the given quadratic equation in the standard form, and identify the values of a, b, and c. ax2+bx+c=0​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. -8x−1=3x2+2LHS+8x=RHS+8x-1=3x2+8x+2LHS+1=RHS+10=3x2+8x+3Rearrange equation3x2+8x+3=0 Now, when the given equation in written in the standard form, we can identify the values of a, b, and c. 3x2+8x+3=0​ We see that a=3, b=8, and c=3.
Exercises 9 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2−12x+36=0⇔1x2+(-12)x+36=0​ We see that a=1, b=-12, and c=36. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-12)±(-12)2−4(1)(36)​​ Solve for x and Simplify -(-a)=ax=2(1)12±(-12)2−4(1)(36)​​Calculate powerx=2(1)12±144−4(1)(36)​​Identity Property of Multiplicationx=212±144−4(36)​​Multiplyx=212±144−144​​Subtract termx=212±0​​Calculate root x=212±0​ Since adding or subtracting zero doesn't change the value of a number, the numerator will simplify to 12. Therefore, we will get only one value of x. x=212​⇔x=6​ Using the Quadratic Formula, we found that the solution of the given equation is x=6.
Exercises 10 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2+7x+16=0⇔1x2+7x+16=0​ We see that a=1, b=7, and c=16. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-7±72−4(1)(16)​​ Solve for x and Simplify Calculate powerx=2(1)-7±49−4(1)(16)​​Identity Property of Multiplicationx=2-7±49−4(16)​​Multiplyx=2-7±49−64​​Subtract term x=2-7±-15​​ Since we cannot calculate the root of a negative number, there are no real solutions of the given equation.
Exercises 11 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2−10x−11=0⇔1x2+(-10)x+(-11)=0​ We see that a=1, b=-10, and c=-11. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-10)±(-10)2−4(1)(-11)​​ Solve for x and Simplify -(-a)=ax=2(1)10±(-10)2−4(1)(-11)​​(-a)2=a2x=2(1)10±100−4(1)(-11)​​Identity Property of Multiplicationx=210±100−4(-11)​​-a(-b)=a⋅bx=210±100+44​​Subtract termx=210±144​​Calculate rootx=210±12​Factor out 2x=22(5±6)​Cancel out common factors x=5±6 The solutions for this equation are x=5±6. Let's separate them into the positive and negative cases.x=5±6 x=5+6x=5−6 x=11x=-1 Using the Quadratic Formula, we found that the solutions of the given equation are x=11 and x=-1.
Exercises 12 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. 2x2−1x−1=0⇔2x2+(-1)x+(-1)=0​ We see that a=2, b=-1, and c=-1. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(2)-(-1)±(-1)2−4(2)(-1)​​ Solve for x and Simplify -(-a)=ax=2(2)1±(-1)2−4(2)(-1)​​(-a)2=a2x=2(2)1±1−4(2)(-1)​​Multiplyx=41±1−8(-1)​​-a(-b)=a⋅bx=41±1+8​​Add termsx=41±9​​Calculate root x=41±3​ The solutions for this equation are x=41±3​. Let's separate them into the positive and negative cases.x=41±3​ x=41+3​x=41−3​ x=44​x=4-2​ x=1x=-21​ Using the Quadratic Formula, we found that the solutions of the given equation are x=1 and x=-21​.
Exercises 13 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. 2x2−6x+5=0⇔2x2+(-6)x+5=0​ We see that a=2, b=-6, and c=5. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(2)-(-6)±(-6)2−4(2)(5)​​ Solve for x and Simplify -(-a)=ax=2(2)6±(-6)2−4(2)(5)​​(-a)2=a2x=2(2)6±36−4(2)(5)​​Multiplyx=46±36−40​​Subtract term x=46±-4​​ Since we cannot calculate the root of a negative number, there are no real solutions of the given equation.
Exercises 14 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. 9x2−6x+1=0⇔9x2+(-6)x+1=0​ We see that a=9, b=-6, and c=1. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(9)-(-6)±(-6)2−4(9)(1)​​ Solve for x and Simplify -(-a)=ax=2(9)6±(-6)2−4(9)(1)​​(-a)2=a2x=2(9)6±36−4(9)(1)​​Multiplyx=186±36−36​​Subtract termx=186±0​​Calculate root x=186±0​ Since adding or subtracting zero doesn't change the value of a number, the numerator will simplify to 6. Therefore, we will get only one value of x. x=186​⇔x=31​​ Using the Quadratic Formula, we found that the solution of the given equation is x=31​.
Exercises 15 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 6x2−13x=-6LHS+6=RHS+66x2−13x+6=0 Now, we can identify the values of a, b, and c. 6x2−13x+6=0⇔6x2+(-13)x+6=0​ We see that a=6, b=-13, and c=6. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(6)-(-13)±(-13)2−4(6)(6)​​ Solve for x and Simplify -(-a)=ax=2(6)13±(-13)2−4(6)(6)​​(-a)2=a2x=2(6)13±169−4(6)(6)​​Multiplyx=1213±169−144​​Subtract termx=1213±25​​Calculate root x=1213±5​ The solutions for this equation are x=1213±5​. Let's separate them into the positive and negative cases.x=1213±5​ x=1213+5​x=1213−5​ x=1218​x=128​ x=23​x=32​ Using the Quadratic Formula, we found that the solutions of the given equation are x=23​ and x=32​.
Exercises 16 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. -3x2+6x=4LHS−4=RHS−4-3x2+6x−4=0 Now, we can identify the values of a, b, and c. -3x2+6x−4=0⇔-3x2+6x+(-4)=0​ We see that a=-3, b=6, and c=-4. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(-3)-6±62−4(-3)(-4)​​ Solve for x and Simplify Calculate powerx=2(-3)-6±36−4(-3)(-4)​​-a(-b)=a⋅bx=2(-3)-6±36+12(-4)​​a(-b)=-a⋅bx=-6-6±36−48​​Add terms x=-6-6±-12​​ Since we cannot calculate the root of a negative number, there are no real solutions of the given equation.
Exercises 17 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 1−8x=-16x2LHS+16x2=RHS+16x216x2+1−8x=0Commutative Property of Addition16x2−8x+1=0 Now, we can identify the values of a, b, and c. 16x2−8x+1=0⇔16x2+(-8)x+1=0​ We see that a=16, b=-8, and c=1. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(16)-(-8)±(-8)2−4(16)(1)​​ Solve for x and Simplify -(-a)=ax=2(16)8±(-8)2−4(16)(1)​​(-a)2=a2x=2(16)8±64−4(16)(1)​​Multiplyx=328±64−64​​Subtract termx=328±0​​Calculate root x=328±0​ Since adding or subtracting zero doesn't change the value of a number, the numerator will simplify to 8. Therefore, we will get only one value of x. x=328​⇔x=41​​ Using the Quadratic Formula, we found that the solution of the given equation is x=41​.
Exercises 18 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2−5x+3=0⇔1x2+(-5)x+3=0​ We see that a=1, b=-5, and c=3. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-5)±(-5)2−4(1)(3)​​ Solve for x and Simplify -(-a)=ax=2(1)5±(-5)2−4(1)(3)​​(-a)2=a2x=2(1)5±25−4(1)(3)​​Identity Property of Multiplicationx=25±25−4(3)​​Multiplyx=25±25−12​​Subtract term x=25±13​​ Using the Quadratic Formula, we found that the solutions of the given equation are x=25±13​​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest tenth.x=25±13​​ x=25+13​​x=25−13​​ x=25+3.6055…​x=25−3.6055…​ x≈25+3.6​x≈25−3.6​ x≈28.6​x≈21.4​ x≈4.3x≈0.7 We found that there are two solutions for the given equation, x≈4.3 and x≈0.7.
Exercises 19 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. x2+2x=9LHS−9=RHS−9x2+2x−9=0 Now, we can identify the values of a, b, and c. x2+2x−9=0⇔1x2+2x+(-9)=0​ We see that a=1, b=2, and c=-9. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-2±22−4(1)(-9)​​ Solve for x and Simplify Calculate powerx=2(1)-2±4−4(1)(-9)​​Identity Property of Multiplicationx=2-2±4−4(-9)​​-a(-b)=a⋅bx=2-2±4+36​​Add termsx=2-2±40​​Split into factorsx=2-2±4⋅10​​a⋅b​=a​⋅b​2-2±4​⋅10​​Calculate root2-2±210​​Factor out 222(-1±10​)​Cancel out common factors x=-1±10​ Using the Quadratic Formula, we found that the solutions of the given equation are x=-1±10​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest tenth.x=-1±10​ x=-1+10​x=-1−10​ x=-1+3.1622…x=-1−3.1622… x≈-1+3.2x≈-1−3.2 x≈2.2x≈-4.2 We found that there are two solutions for the given equation, x≈2.2 and x≈-4.2.
Exercises 20 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 5x2−2=4xLHS−4x=RHS−4x5x2−4x−2=0 Now, we can identify the values of a, b, and c. 5x2−4x−2=0⇔5x2+(-4)x+(-2)=0​ We see that a=5, b=-4, and c=-2. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(5)-(-4)±(-4)2−4(5)(-2)​​ Solve for x and Simplify -(-a)=ax=2(5)4±(-4)2−4(5)(-2)​​(-a)2=a2x=2(5)4±16−4(5)(-2)​​Multiplyx=104±16−20(-2)​​-a(-b)=a⋅bx=104±16+40​​Add termsx=104±56​​Split into factorsx=104±4⋅14​​a⋅b​=a​⋅b​x=104±4​⋅14​​Calculate rootx=104±214​​Factor out 2x=102(2±14​)​ba​=b/2a/2​ x=52±14​​ Using the Quadratic Formula, we found that the solutions of the given equation are x=52±14​​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest tenth.x=52±14​​ x=52+14​​x=52−14​​ x=52+3.7416…​x=52−3.7416…​ x≈52+3.7​x≈52−3.7​ x≈55.7​x≈5-1.7​ x≈1.1x≈-0.3 We found that there are two solutions for the given equation, x≈1.1 and x≈-0.3.
Exercises 21 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 2x2+9x+7=3LHS−3=RHS−32x2+9x+4=0 Now, we can identify the values of a, b, and c. 2x2+9x+4=0​ We see that a=2, b=9, and c=4. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(2)-9±92−4(2)(4)​​ Solve for x and Simplify Calculate powerx=2(2)-9±81−4(2)(4)​​Multiplyx=4-9±81−32​​Subtract termx=4-9±49​​Calculate root x=4-9±7​ The solutions for this equation are x=4-9±7​. Let's separate them into the positive and negative cases.x=4-9±7​ x=4-9+7​x=4-9−7​ x=4-2​x=4-16​ x=-21​x=-4 Using the Quadratic Formula, we found that the solutions of the given equation are x=-21​ and x=-4.
Exercises 22 We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 8x2+8=6−9xLHS−6=RHS−68x2+2=-9xLHS+9x=RHS+9x8x2+9x+2=0 Now, we can identify the values of a, b, and c. 8x2+9x+2=0​ We see that a=8, b=9, and c=2. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(8)-9±92−4(8)(2)​​ Solve for x and Simplify -(-a)=ax=2(8)-9±92−4(8)(2)​​(-a)2=a2x=2(8)-9±81−4(8)(2)​​Multiplyx=16-9±81−64​​Add terms x=16-9±17​​ Using the Quadratic Formula, we found that the solutions of the given equation are x=16-9±17​​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest tenth.x=16-9±17​​ x=16-9+17​​x=16-9−17​​ x=16-9+4.1231…​x=16-9−4.1231…​ x≈5-9+4.1​x≈5-9−4.1​ x≈5-4.9​x≈5-13.1​ x≈-1.0x≈-2.6 We found that there are two solutions for the given equation, x≈-1.0 and x≈-2.6.
Exercises 23 In this exercise, we are given a function and definitions for the two variables. Let's identify where the provided values belong by analyzing the function. h=-16t2+26t​ In this model, h represents the height and t represents number of seconds. We are asked at what time does the dolphin reach 5 ft. Let's substitute h=5 and put quadratic in standard form. h=-16t2+26th=55=-16t2+26tLHS−5=RHS−50=-16t2+26t−5Rearrange equation-16t2+26t−5=0 Now, we can apply the Quadratic Formula and solve for t. -16t2+26t−5=0Use the Quadratic Formula: a=-16,b=26,c=-5t=2(-16)-(26)±(26)2−4(-16)(-5)​​ Simplify right-hand side Calculate power and productt=-32-26±676−320​​Add and subtract termst=-32-26±356​​Calculate root t≈-32-26±18.87​ The solutions for this equation are t≈-32-26±18.87​. Let's separate the two cases.t≈-32-26±18.87​ t1​≈t2​≈ -32-26+18.87​-32-26−18.87​ -32-7.13​-32-44.87​ 0.221.402 We can see that the dolphin was at 5 ft after about 0.2 sec and again at about 1.4 seconds.
Exercises 24
Exercises 25 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Notice that the equation is given in the standard form, so we can now identify the values of a, b, and c. x2−6x+10=0⇔1x2+(-6)x+10=0​ Now, let's evaluate the discriminant. b2−4acSubstitute values(-6)2−4(1)(10) Simplify (-a)2=a236−4(1)(10)Identity Property of Multiplication36−4(10)Multiply36−40Subtract term -4 Since the discriminant is -4, the quadratic equation has no real solutions. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21513_0_836825372_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21513_0_836825372_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21513_0_836825372_l", "Solution21513_0_836825372_p", 1, code); }); } ); } window.JXQtable["Solution21513_0_836825372_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1630206294_469137638').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1449360718_501631291').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1612904139_429214134').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1768282932_20001864').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 26 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Notice that the equation is given in the standard form, so we can now identify the values of a, b, and c. x2−5x−3=0⇔1x2+(-5)x+(-3)=0​ Now, let's evaluate the discriminant. b2−4acSubstitute values(-5)2−4(1)(-3) Simplify (-a)2=a225−4(1)(-3)Identity Property of Multiplication25−4(-3)-a(-b)=a⋅b25+12Add terms 37 Since the discriminant is 37, the quadratic equation has two real solutions. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21514_0_1394922097_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21514_0_1394922097_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21514_0_1394922097_l", "Solution21514_0_1394922097_p", 1, code); }); } ); } window.JXQtable["Solution21514_0_1394922097_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1172182161_106229029').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1539888281_2114232857').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn970473947_1664637245').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn280426338_1306567637').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 27 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first rewrite the given equation in standard form. 2x2−12x=-18LHS+18=RHS+182x2−12x+18=0 Having rewritten the equation, we can now identify the values of a, b, and c. 2x2−12x+18=0⇔2x2+(-12)x+18=0​ Finally, let's evaluate the discriminant. b2−4acSubstitute values(-12)2−4(2)(18) Simplify (-a)2=a2144−4(2)(18)Multiply144−144Subtract term 0 Since the discriminant is 0, the quadratic equation has one real solution. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21515_0_1524760380_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21515_0_1524760380_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21515_0_1524760380_l", "Solution21515_0_1524760380_p", 1, code); }); } ); } window.JXQtable["Solution21515_0_1524760380_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn632441019_259205960').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1311094660_2076551823').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1999726410_370634695').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1148370098_319147986').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 28 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first rewrite the given equation in standard form. 4x2=4x−1LHS−4x=RHS−4x4x2−4x=-1LHS+1=RHS+14x2−4x+1=0 Having rewritten the equation, we can now identify the values of a, b, and c. 4x2−4x+1=0⇔4x2+(-4)x+1=0​ Finally, let's evaluate the discriminant. b2−4acSubstitute values(-4)2−4(4)(1) Simplify (-a)2=a216−4(4)(1)Identity Property of Multiplication16−4(4)Multiply16−16Subtract term 0 Since the discriminant is 0, the quadratic equation has one real solution. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21516_0_961479701_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21516_0_961479701_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21516_0_961479701_l", "Solution21516_0_961479701_p", 1, code); }); } ); } window.JXQtable["Solution21516_0_961479701_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1890827323_315883101').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn626838509_1848005690').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1999334216_167384982').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn744016608_217928162').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 29 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first rewrite the given equation in standard form. -41​x2+4x=-2LHS+2=RHS+2-41​x2+4x+2=0 Having rewritten the equation, we can now identify the values of a, b, and c. -41​x2+4x+2=0⇔-41​x2+4x+2=0​ Finally, let's evaluate the discriminant. b2−4acSubstitute values42−4(-41​)(2) Simplify Calculate power16−4(-41​)(2)-a(-b)=a⋅b16+4(41​)(2)a⋅b1​=ba​16+44​(2)Calculate quotient16−1(2)Identity Property of Multiplication16−2Subtract term 14 Since the discriminant is 14, the quadratic equation has two real solutions. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21517_0_2143830811_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21517_0_2143830811_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21517_0_2143830811_l", "Solution21517_0_2143830811_p", 1, code); }); } ); } window.JXQtable["Solution21517_0_2143830811_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn518821505_421061184').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn539609938_170062124').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn383137310_991410772').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn948589427_805570042').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 30 We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first rewrite the given equation in standard form. -5x2+8x=9LHS−9=RHS−9-5x2+8x−9=0 Having rewritten the equation, we can now identify the values of a, b, and c. -5x2+8x−9=0⇔-5x2+8x+(-9)=0​ Finally, let's evaluate the discriminant. b2−4acSubstitute values82−4(-5)(-9) Simplify Calculate power64−4(-5)(-9)-a(-b)=a⋅b64+20(-9)a(-b)=-a⋅b64−180Subtract term -116 Since the discriminant is -116, the quadratic equation has no real solutions. Extra info Further information If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21518_0_142962486_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21518_0_142962486_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21518_0_142962486_l", "Solution21518_0_142962486_p", 1, code); }); } ); } window.JXQtable["Solution21518_0_142962486_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn180450887_486595248').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn17980861_2043938049').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1966568258_1795278272').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn662072657_158727774').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 31 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. y=x2+5x−1⇔y=1x2+5x+(-1)​ Now, let's evaluate the discriminant. b2−4acSubstitute values52−4(1)(-1) Simplify Calculate power25−4(1)(-1)Identity Property of Multiplication25−4(-1)-a(-b)=a⋅b25+4Add terms 29 Since the discriminant is 29, the quadratic equation has two x-intercepts. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21519_0_1070981162_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21519_0_1070981162_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21519_0_1070981162_l", "Solution21519_0_1070981162_p", 1, code); }); } ); } window.JXQtable["Solution21519_0_1070981162_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1734875677_790313091').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn416338479_1844066011').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn629592352_894302716').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1786125436_1488681638').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 32 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. y=4x2+4x+1⇔y=4x2+4x+1​ Now, let's evaluate the discriminant. b2−4acSubstitute values42−4(4)(1) Simplify Calculate power16−4(4)(1)Identity Property of Multiplication16−4(4)Multiply16−16Subtract term 0 Since the discriminant is 0, the quadratic equation has one x-intercept. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21520_0_833539162_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21520_0_833539162_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21520_0_833539162_l", "Solution21520_0_833539162_p", 1, code); }); } ); } window.JXQtable["Solution21520_0_833539162_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1715578286_539521803').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1173793663_1516088771').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1556000032_651382794').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1510373606_589033114').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 33 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. y=-6x2+3x−4⇔y=-6x2+3x+(-4)​ Now, let's evaluate the discriminant. b2−4acSubstitute values32−4(-6)(-4) Simplify Calculate power9−4(-6)(-4)-a(-b)=a⋅b9+24(-4)Multiply9−96Add terms -87 Since the discriminant is -87, the quadratic equation has no x-intercepts. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21521_0_2050313611_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21521_0_2050313611_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21521_0_2050313611_l", "Solution21521_0_2050313611_p", 1, code); }); } ); } window.JXQtable["Solution21521_0_2050313611_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn322232824_225434730').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn536623663_1022559350').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1574024636_2095138808').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn904580457_1582641730').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 34 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. y=-1x2+5x+13​ Now, let's evaluate the discriminant. b2−4acSubstitute values52−4(-1)(13) Simplify Calculate power25−4(-1)(13)-a(-b)=a⋅b25+4(13)Multiply25+52Add terms 77 Since the discriminant is 77, the quadratic equation has two x-intercepts. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21522_0_1201745533_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21522_0_1201745533_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21522_0_1201745533_l", "Solution21522_0_1201745533_p", 1, code); }); } ); } window.JXQtable["Solution21522_0_1201745533_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2124062223_1496063673').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn931035350_1319612780').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn188691908_2127117737').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn162585721_359657693').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 35 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. f(x)=4x2+3x−6⇔f(x)=4x2+3x+(-6)​ Now, let's evaluate the discriminant. b2−4acSubstitute values32−4(4)(-6) Simplify Calculate power9−4(4)(-6)Multiply9−16(-6)-a(-b)=a⋅b9+96Add terms 105 Since the discriminant is 105, the quadratic equation has two x-intercepts. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21523_0_1876042737_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21523_0_1876042737_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21523_0_1876042737_l", "Solution21523_0_1876042737_p", 1, code); }); } ); } window.JXQtable["Solution21523_0_1876042737_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2006006859_542454631').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn655091994_1522136857').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1580715059_983487623').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1190972620_554026703').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 36 We want to use the discriminant of the given quadratic function to determine the number of x-intercepts. In the Quadratic Formula, b2−4ac is the discriminant. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ If we just want to know the number of x-intercepts, and not the x-intercepts themselves, we only need to work with the discriminant. We can now identify the values of a, b, and c. f(x)=2x2+8x+8⇔f(x)=2x2+8x+8​ Now, let's evaluate the discriminant. b2−4acSubstitute values82−4(2)(8) Simplify Calculate power64−4(2)(8)Multiply64−64Subtract term 0 Since the discriminant is 0, the quadratic equation has one x-intercept. Extra info Further information If the discriminant is greater than zero, the equation will have two x-intercepts. If it is equal to zero, the equation will have one x-intercept. Finally, if the discriminant is less than zero, the equation will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21524_0_2015016862_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {}; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); //A Two real solutions: blue #628aff and #628aff mlg.af("showA",function() { for(var item in el){ b.remove(el[item]); } console.log(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xmax:1.7,xmin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#628aff", fontsize:0.8 } ); }); /*B One Real Solution: green #00c997 and #00c997*/ mlg.af("showB",function() { for(var item in el){ b.remove(el[item]); } console.log(b.board); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#00c997", fontsize:0.8 } ); }); //C No Real Solutions: orange #ff8a62 and #ff8a62 ff8a62 mlg.af("showC",function() { for(var item in el){ b.remove(el[item]); } el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff8a62'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); }); //D Reset mlg.af("showD",function() { for(var item in el){ b.remove(el[item]); } el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{Try these values}\\\\\\text{of the discriminant!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:1000 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21524_0_2015016862_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21524_0_2015016862_l", "Solution21524_0_2015016862_p", 1, code); }); } ); } window.JXQtable["Solution21524_0_2015016862_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2011463092_1073775744').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );=0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1886789546_641394648').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );<0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1959254575_1713106611').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2080315416_1505175137').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 37 The given quadratic equation is not easily factorable, therefore we will solve it using the Quadratic Formula. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. -10x2+13x=4LHS−4=RHS−4-10x2+13x−4=0 Now, we can identify the values of a, b, and c. -10x2+13x−4=0⇔-10x2+13x+(-4)=0​ We see that a=-10, b=13, and c=-4. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(-10)-13±132−4(-10)(-4)​​ Solve for x and Simplify Calculate powerx=2(-10)-13±169−4(-10)(-4)​​-a(-b)=a⋅bx=2(-10)-13±169+40(-4)​​a(-b)=-a⋅bx=-20-13±169−160​​Subtract termx=-20-13±9​​Calculate rootx=-20-13±3​ba​=b/(-1)a/(-1)​ x=2013±3​ The solutions for this equation are x=2013±3​. Let's separate them into the positive and negative cases.x=2013±3​ x=2013+3​x=2013−3​ x=2016​x=2010​ x=54​x=21​ Using the Quadratic Formula, we found that the solutions of the given equation are x=54​ and x=21​.
Exercises 38 The given quadratic equation is not easily factorable, therefore we will solve it using the Quadratic Formula. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2−3x−40=0⇔1x2+(-3)x+(-40)=0​ We see that a=1, b=-3, and c=-40. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-3)±(-3)2−4(1)(-40)​​ Solve for x and Simplify -(-a)=ax=2(1)3±(-3)2−4(1)(-40)​​(-a)2=a2x=2(1)3±9−4(1)(-40)​​Identity Property of Multiplicationx=23±9−4(-40)​​-a(-b)=a⋅bx=23±9+160​​Add termsx=23±169​​Calculate root x=23±13​ The solutions for this equation are x=23±13​. Let's separate them into the positive and negative cases.x=23±13​ x=23+13​x=23−13​ x=216​x=2-10​ x=8x=-5 Using the Quadratic Formula, we found that the solutions of the given equation are x=8 and x=-5.
Exercises 39 The given quadratic equation is not easily factorable, but a=1 and b is even, therefore we will solve it by Completing the Square. Note that all terms with x are on one side of the equation. x2+6x=5​ In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=2. Let's now calculate (2b​)2. (2b​)2b=6(26​)2Calculate quotient32Calculate power9 Next, we will add (2b​)2=9 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation. If you need further explanations on how to factor this kind of expression, please see this exercise or this exercise. x2+6x=5LHS+9=RHS+9x2+6x+9=5+9a2+2ab+b2=(a+b)2(x+3)2=5+9Add terms(x+3)2=14LHS​=RHS​(x+3)2​=14​Calculate rootx+3=±14​LHS−1=RHS−1x=-3±14​ The solutions for this equation are x=-3±14​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest hundredth.x=-3±14​ x=-3+14​x=-3−14​ x=-3+3.7416…x=-3−3.7416… x≈-3+3.74x≈-3−3.74 x≈0.74x≈-6.74 We found that there are two solutions for the given equation, x≈0.74 and x≈-6.74.
Exercises 40 Notice that there is no x-term, only the x2-term and constants, so we can rewrite the given equation to obtain the form x2=d. Then we will solve the equation by taking the square roots. Remember that we need to consider the positive and negative solutions. -5x2=-25LHS/(-5)=RHS/(-5)x2=5LHS​=RHS​x2​=5​Calculate rootx=±5​Use a calculatorx≈±2.2360…x≈±2.24 We found that x≈±2.24. Thus, there are two solutions for the equation: x≈2.24 and x≈-2.24.
Exercises 41 The given equation is easily factorable, so we will solve it by factoring.Factoring Let's start by rewriting the x-term as a difference. x2+x−12=0Write as a differencex2+4x−3x−12=0 Factor out x & -3 Factor out xx(x+4)−3x−12=0Factor out -3 x(x+4)−3(x+4)=0Factor out (x+4)(x−3)(x+4)=0 For further explanation on how to factor this type of expression, please see this example.Solving To solve this equation, we will apply the Zero Product Property. (x−3)(x+4)=0Use the Zero Product Propertyx−3=0x+4=0​(I)(II)​(I):  LHS+3=RHS+3x=3x+4=0​(II):  LHS−4=RHS−4x=3x=-4​ We found that the solutions of the given equation are x=3 and x=-4.
Exercises 42 The given quadratic equation is not easily factorable, therefore we will solve it using the Quadratic Formula. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ We first need to identify the values of a, b, and c. x2−4x+1=0⇔1x2+(-4)x+1=0​ We see that a=1, b=-4, and c=1. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-4)±(-4)2−4(1)(1)​​ Solve for x and Simplify -(-a)=ax=2(1)4±(-4)2−4(1)(1)​​(-a)2=a2x=2(1)4±16−4(1)(1)​​Identity Property of Multiplicationx=24±16−4​​Subtract termx=24±12​​Split into factorsx=24±4⋅3​​a⋅b​=a​⋅b​x=24±4​⋅3​​Calculate rootx=24±23​​Factor out 2x=22(2±3​)​Cancel out common factors x=2±3​ Using the Quadratic Formula, we found that the solutions of the given equation are x=2±3​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest hundredth.x=2±3​ x=2+3​x=2−3​ x=2+1.7320…x=2−1.7320… x≈2+1.73x≈2−1.73 x≈3.73x≈0.27 We found that there are two solutions for the given equation, x≈3.73 and x≈0.27.
Exercises 43 The given quadratic equation is not easily factorable, therefore we will solve it using the Quadratic Formula. ax2+bx+c=0⇔x=2a-b±b2−4ac​​​ Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible. 4x2−x=17LHS−17=RHS−174x2−x−17=0 Now, we can identify the values of a, b, and c. 4x2−x−17=0⇔4x2+-1x+(-17)=0​ We see that a=4, b=-1, and c=-17. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(4)-(-1)±(-1)2−4(4)(-17)​​ Solve for x and Simplify -(-a)=ax=2(4)1±(-1)2−4(4)(-17)​​(-a)2=a2x=2(4)1±1−4(4)(-17)​​Multiplyx=81±1−16(-17)​​-a(-b)=a⋅bx=81±1+272​​Add terms x=81±273​​ Using the Quadratic Formula, we found that the solutions of the given equation are x=81±273​​. Now, let's separate them by using the positive and negative signs. Then we will calculate the value of the square root and round it to the nearest tenth.x=81±273​​ x=81+273​​x=81−273​​ x=81+16.5227…​x=81−16.5227…​ x≈81+16.52​x≈81−16.52​ x≈817.52​x≈8-15.52​ x≈2.19x≈-1.94 We found that there are two solutions for the given equation, x≈2.19 and x≈-1.94.
Exercises 44 The given equation is easily factorable, so we will solve it by factoring.Factoring Let's start by writing all the terms on one side of the equals sign. x2+6x+9=16LHS−16=RHS−16x2+6x−7=0Write as a differencex2+7x−x−7=0 Factor out x & -1 Factor out xx(x+7)−x−7=0Factor out -1 x(x+7)−(x+7)=0Factor out (x+7)(x−1)(x+7)=0 For further explanation on how to factor this type of expression, please see this example.Solving To solve this equation, we will apply the Zero Product Property. (x−1)(x+7)=0Use the Zero Product Propertyx−1=0x+7=0​(I)(II)​(I):  LHS+1=RHS+1x=1x+7=0​(II):  LHS−7=RHS−7x=1x=-7​ We found that the solutions of the given equation are x=1 and x=-7.
Exercises 45 The Quadratic Formula has many computations and, therefore, many places where errors can occur. In this case we can see that the first time b=-7 was replaced in the Quadratic Formula, one of the negatives was dropped. x​=​2a-b±b2−4ac​​​ In our quadratic, 3x2−7x−6=0, we have a=3, b=-7, and c=-6. So the first substitution of -b should have looked like -(-7).Let's substitute correctly and solve.Interestingly enough, the answers with the error are just different signs of the actual solutions.
Exercises 46 When using the Quadratic Formula, there are many computations and, therefore, many opportunities to introduce errors. In this case, we can see the error came before the substitution started. -2x2+9x=4​ The equation is not in standard form. Let's first put the equation into standard form by subtracting 4 from both sides. -2x2+9x−4=0​ To use the Quadratic Formula correctly we need to substitute a=-2, b=9, and c=-4. Let's look to see if there is an error in substitution.We can see the error, 4 instead of -4 was used for c. Let's correct the error.Alternative solution info Step by Step Solution -2x2+9x−4=0Use the Quadratic Formula: a=-2,b=9,c=-4x=2(-2)-(9)±(9)2−4(-2)(-4)​​ Simplify right-hand side Calculate power and productx=-4-9±81−32​​Add and subtract termsx=-4-9±49​​Calculate root x=-4-9±7​ The solutions for this equation are x=-4-9±7​. Let's separate the two cases.x=-4-9±7​ x1​=-4-9+7​x2​=-4-9−7​ x1​=-4-2​x2​=-4-16​ x1​=21​x2​=4
Exercises 47 In this exercise, we are given an equation with two variables. Let's look to see what the variables represent then substitute and solve the equation. The first question asks if the water can reach a height of 50 feet. Since y represents height, let's start by substituting y=50 into the quadratic, then put it in standard form to solve. y=-0.006x2+1.2x+10y=5050=-0.006x2+1.2x+10LHS−50=RHS−500=-0.006x2+1.2x−40Rearrange equation-0.006x2+1.2x−40=0 Now that the equation is in standard form, we can apply the Quadratic Formula to solve for x. -0.006x2+1.2x−40=0Use the Quadratic Formula: a=-0.006,b=1.2,c=-40x=2⋅-0.006-(1.2)±(1.2)2−4⋅-0.006(-40)​​ Simplify right-hand side Calculate power and productx=-0.012-1.2±1.44−0.96​​Add and subtract termsx=-0.012-1.2±0.48​​Calculate root x≈-0.012-1.2±0.693​ The solutions for this equation are x≈-0.012-1.2±0.693​. Let's separate the two cases.x≈-0.012-1.2±0.693​ x1​≈x2​≈ -0.012-1.2+0.693​-0.012-1.2−0.693​ -0.012-0.507​-0.012-1.893​ 42.25157.75 In this exercise, x represents the distance from the river's shore. Our solution shows us that the fountain reaches 50 feet twice, once about 42 feet from the shore and the other at about 158 feet from the shore.
Exercises 48
Exercises 49 While it is possible to use the discriminant to determine the number of solutions, we do not know if our friend computed it correctly. To compute the discriminant, we need to first put the equation in standard form and identify a, b, and c. 2x2−5x−2=-11LHS+11=RHS+112x2−5x−2+11=-11+11Add and subtract terms2x2−5x−9=0 Now that our equation is in standard form we can identify our standard coefficients a=2, b=-5, and c=-9. Let's look at the formula for the discriminant and substitute our values. d↓d​==​b2↓(-5)2​−−​4↓4​⋅⋅​a↓2​⋅⋅​c↓-9​ We can compute the discriminant and get d=97. Since that is greater than zero, the quadratic will have two real solutions. Therefore, our friend is correct.
Exercises 50 In this exercise, we are given an equation with two variables and a situation about a tent. The question asks if it's possible for a person 4 feet could stand in the tent. Let's have a look at the curve.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21538_1_601660406_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b=mlg.board([-0.5,4.5,9.5,-0.5],{"desktopSize":"medium","style":"usa"});/*default display*/ b.func("-0.18*x^2+1.6*x"); b.func("4");/*axes*/ b.xaxis(10,9); b.yaxis(10,9);} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21538_1_601660406_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21538_1_601660406_l", "Solution21538_1_601660406_p", 1, code); }); } ); } window.JXQtable["Solution21538_1_601660406_l"] = true;The line above the curve is at y=4 and we can see by the picture the tent lies completely below it. But to solve this without a picture, we only need to determine if the discriminant is greater than or equal to zero when the height of the arc is 4 feet. Let's substitute y=4, put the equation in standard form, then find the discriminant. y=−0.18x2+1.6xy=44=−0.18x2+1.6xLHS−4=RHS−40=−0.18x2+1.6x−4Rearrange equation−0.18x2+1.6x−4=0 With the equation in standard form, we have a=-0.18, b=1.6, and c=-4. Let's substitute those into the discriminat formla and compute. d=b2−4acSubstitute valuesd=1.62−4(-0.18)(-4)Calculate power and productd=2.56−2.88Add and subtract termsd=-0.32 Since the discriminant is negative, there are no real solutions. Therefore, we can say that a 4 foot child could not stand up in the tent.
Exercises 51
Exercises 52
Exercises 53
Exercises 54
Exercises 55
Exercises 56
Exercises 57
Exercises 58
Exercises 59
Exercises 60
Exercises 61
Exercises 62
Exercises 63
Exercises 64
Exercises 65
Exercises 66
Exercises 67
Exercises 68
Exercises 69
Exercises 70
Exercises 71
Exercises 72
Exercises 73
Exercises 74
Exercises 75
Exercises 76
Exercises 77
Exercises 78
Exercises 79
Exercises 80
Exercises 81
Exercises 82
Exercises 83
Exercises 84
Exercises 85
Exercises 86