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###### Exercises

Exercise name | Free? |
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Exercises 1 We will solve the given equation by taking the square roots, because it has the form of ax2+c=0. Therefore, we will easily isolate x2 on one side of the equation. Remember we need to consider the positive and negative solutions. x2−121=0LHS+121=RHS+121x2=121LHS=RHSx=±121Calculate rootx=±11 We found that x=±11. Thus, there are two solutions for the equation: x=11 and x=-11. Checking Our Answer info Checking our answer We can check our answers by substituting them for x in the given equation. Let's start with x=-11. x2−121=0x=-11(-11)2−121=?0 Simplify (-a)2=a2(11)2−121=?0Calculate power121−121=?0Subtract term 0=0 ✓ Since 0=0, we know that x=-11 is a solution of the equation. Let's check if x=11 is also a solution. x2−121=0x=11(11)2−121=?0 Simplify Calculate power121−121=?0Subtract term 0=0 ✓ Again, since 0=0, we know that x=11 is a solution of the equation. | |

Exercises 2 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. x2−6x=10⇔x2−6x−10=0 Now, we can identify the values of a, b, and c. x2−6x−10=0⇕(1)x2+(-6)x+(-10)=0 We see that a=1, b=-6, and c=-10. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4acSubstitute valuesx=2(1)-(-6)±(-6)2−4(1)(-10) Solve for x and Simplify -(-a)=ax=2(1)6±(-6)2−4(1)(-10)Identity Property of Multiplicationx=26±(-6)2−4(-10)(-a)2=a2x=26±62−4(-10)Calculate powerx=26±36−4(-10)-a(-b)=a⋅bx=26±36+40Add termsx=24±76Split into factorsx=24±4⋅19a⋅b=a⋅bx=26±4⋅19Calculate rootx=26±219Factor out 2x=22(3±19)Cancel out common factors x=3±19 Using the Quadratic Formula, we found that the solutions of the given equation are x=3±19. Therefore, the solutions are x1=3+19 and x2=3−19. | |

Exercises 3 Since our given quadratic equation is in standard form, we'll use the Quadratic Formula to solve it. ax2+bx+c=0⇔x=2a-b±b2−4ac We first need to identify the values of a, b, and c in the given equation. -2x2+3x+7=0⇔(-2)x2+3x+7=0 We see that a=-2, b=3, and c=7. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4acSubstitute valuesx=2(-2)-3±32−4(-2)(7) Solve for x and Simplify -a(-b)=a⋅bx=2(-2)-3±32+8(7)a(-b)=-a⋅bx=-4-3±32+8(7)Calculate powerx=-4-3±9+8(7)Multiplyx=-4-3±9+56Add termsx=-4-3±65Factor out -1 x=43±65 Using the Quadratic Formula, we found that the solutions of the given equation are x=43±65. Therefore, the solutions are x1=43+65 and x2=43−65. | |

Exercises 4 Since our given quadratic equation is in standard form, we'll use the Quadratic Formula to solve it. ax2+bx+c=0⇔x=2a-b±b2−4ac We first need to identify the values of a, b, and c in the given equation. x2−7x+12=0⇔(1)x2+(-7)x+12=0 We see that a=1, b=-7, and c=12. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4acSubstitute valuesx=2(1)-(-7)±(-7)2−4(1)(12) Solve for x and Simplify -(-a)=ax=2(1)7±(-7)2−4(1)(12)Identity Property of Multiplicationx=27±(-7)2−4(12)(-a)2=a2x=27±72−4(12)Calculate powerx=27±49−4(12)Multiplyx=27±49−48Subtract termx=27±1Calculate root x=27±1 The solutions for this equation are x=27±1. Let's separate them into the positive and negative cases.x=27±1 x1=27+1x2=27−1 x1=28x2=26 x1=4x2=3 Using the Quadratic Formula, we found that the solutions of the given equation are x1=4 and x2=3. | |

Exercises 5 Since our given quadratic equation is in standard form, we'll use the Quadratic Formula to solve it. ax2+bx+c=0⇔x=2a-b±b2−4ac We first need to identify the values of a, b, and c in the given equation. 5x2+x−4=0⇔5x2+(1)x+(-4)=0 We see that a=5, b=1, and c=-4. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4acSubstitute valuesx=2(5)-1±(1)2−4(5)(-4) Solve for x and Simplify 1a=1x=2(5)-1±1−4(5)(-4)Multiplyx=10-1±1−20(-4)-a(-b)=a⋅bx=10-1±1+80Add termsx=10-1±81Calculate root x=10-1±9 The solutions for this equation are x=10-1±9. Let's separate them into the positive and negative cases.x=10-1±9 x1=10-1+9x2=10-1−9 x1=108x2=10-10 x1=54x2=-1 Using the Quadratic Formula, we found that the solutions of the given equation are x1=54 and x2=-1. | |

Exercises 6 To solve the given equation by taking the square roots. Each side of the given equation is a square, therefore we can solve it by taking square root of each side. (4x+3)2=16LHS=RHS4x+3=±16Calculate root4x+3=±4LHS−3=RHS−34x=±4−3LHS/4=RHS/4x=4±4−3The solutions for this equation are x=4±4−3. Let's separate them into the positive and negative cases.x=4±4−3 x1=44−3x2=4-4−3 x1=41x2=-47Therefore, there are two solutions for the equation: x=41 and x=-47. Checking Our Answer info Checking our answer We can check our answers by substituting them for x in the given equation. Let's start with x=41. (4x+3)2=16x=41(4(41)+3)2=?16 Simplify a⋅b1=ba(44+3)2=?16aa=1(1+3)2=?16Add terms(4)2=?16Calculate power 16=16 ✓ Since 16=16, we know that x=41 is a solution of the equation. Let's check if x=-47 is also a solution. (4x+3)2=16x=-47(4(-47)+3)2=?16 Simplify a⋅cb=ca⋅b(-44(7)+3)2=?16Multiply(-428+3)2=?16Calculate quotient(-7+3)2=?16Add terms(-4)2=?16(-a)2=a2(4)2=?16Calculate power 16=16 ✓ Again, since 16=16, we know that x=-47 is a solution of the equation. | |

Exercises 7 Recall that by completing the square we can write a quadratic function in vertex form. f(x)=a(x−h)2+k In this form, a gives the direction of the parabola. When a>0, the parabola opens up, and when a<0, it opens down. The vertex of the parabola lies at (h,k). These characteristics can help us to determine if the exercise's graph corresponds to the function. We will show how to do this. Let's take a look at the given function. f(x)=2x2+4x−6 Notice that the function has a greatest common factor of 2. The first thing to do in order to complete the square is to factor out 2. 2x2+4x−6⇔2(x2+2x−3) Now that we have an expression of the form x2+bx+c inside parenthesis, we can add and subtract (2b)2 to obtain a perfect square trinomial. Let's proceed to identifying b. x2+bx+c2(x2+2x−3) As we can see, b=2. Now we can calculate (2b)2. (2b)2b=2(22)2Calculate quotient(1)21a=11 Therefore, by adding and subtracting 1 to the quantity inside the parenthesis, we can complete the square and rewrite the function in vertex form. f(x)=2(x2+2x−3) Write in vertex form a=a+1−1f(x)=2(x2+2x+1−1−3)a2+2ab+b2=(a+b)2f(x)=2((x+1)2−1−3)Subtract termf(x)=2((x+1)2−4)Distribute 2f(x)=2(x+1)2−2(4)Multiply f(x)=2(x+1)2−8 Now that we have our function written in vertex form, we can identify the parabola's vertex and direction. f(x)=a(x−h)2+kf(x)=2(x−(-1))2+(-8) Because a=2>0, the parabola opens up. Furthermore, we can see that its vertex will be located at (-1,-8), This means that it should lay in quadrant III. By taking a look at the exercise's graph, we can see that it does not correspond to this function, since the graph shows a parabola with a vertex in quadrant IV instead. | |

Exercises 8 This exercise has the freedom to create, which is great. In order to do so, let's first review what a conjugate is. When we use a conjugate, we want two multiply to binomials so the middle term disappears and we have only square terms. Recall the difference of two perfect squares. (a+b)(a−b)=a2−b2 There are many applications for conjugates. But in this case, the exercise wants us to use one with a radical. Let's start by making either a or b a square root, a=10, and b=2. a+b=10+2 Now, if we want it to be used to simplify an expression, we need to put the binomial with the values for a and b in the denominator of a fraction. 10+215 Generally speaking, mathematicians do not like radicals in the denominators of fractions. Therefore, we use the conjugate to simplify the fraction. In this case, our conjugate is 10−2. 10+215 Simplify denominator ba=b⋅(10−2)a⋅(10−2)(10+2)(10−2)15(10−2)Distribute (10−2)(10)2−221510−15⋅2Calculate power and product10−41510−30Add and subtract terms61510−30Simplify quotient 2.510−5 This is a specific example, and the exercise does not require that we simplify it. As long as our expression is in the following form, and our square root number, a, is not a perfect square, we can use any values for a, b, and c. a±bc We can also choose either plus or minus. Also, b and c do not have to be rational. There are many solutions to this exercise. | |

Exercises 9 To solve a system of linear equations using the elimination method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This means that either the x-terms or the y-terms must cancel each other out. {y=x2−4x−2y=-4x+2(I)(II) We can see that the y-terms will eliminate each other if we subtract (II) from (I). {y=x2−4x−2y=-4x+2(I): Subtract (II){y−(y)=x2−4x−2−(-4x+2)y=-4x+2 (II): Solve for x (I): Remove parentheses and change signs{y−y=x2−4x−2+4x−2y=-4x+2(I): Add and subtract terms{0=x2−4y=-4x+2(I): LHS+4=RHS+4{4=x2y=-4x+2(I): \Rearrange equation{x2=4y=-4x+2(I): LHS=RHS{x=±4y=-4x+2Calculate root {x=±2y=-4x+2 Now we can solve for y by substituting values of x into either equation and simplifying. We will use the second equation for the substitution. Notice, we need to consider the positive and negative solutions.xy=-4x+2Solution 2y=-4(2)+2{x=2y=-6 -2y=-4(-2)+2{x=-2y=10 There are two solutions of the system of equations (2,-6) and (-2,10). | |

Exercises 10 We want to solve the given system of equations using the Substitution Method. {y=-5x2+x−1y=-7(I)(II) The y-variable is isolated in Equation (II). This allows us to substitute its value -7 for y in Equation (I). {y=-5x2+x−1y=-7(I): y=-7{-7=-5x2+x−1y=-7 (I): Simplify (I): LHS+7=RHS+7{0=-5x2+x−1+7y=-7(I): Add terms{0=-5x2+x+6y=-7(I): Rearrange equation {-5x2+x+6=0y=-7 Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. -5x2+x+6=0⇔(-5)x2+(1)x+6=0 Now, recall the Quadratic Formula. x=2a-b±b2−4ac We can substitute a=-5, b=1, and c=6 into this formula to solve the quadratic equation. x=2a-b±b2−4acSubstitute valuesx=2(-5)-1±(1)2−4(-5)(6) Solve for x 1a=1x=2(-5)-1±1−4(-5)(6)-a(-b)=a⋅bx=2(-5)-1±1+20(6)a(-b)=-a⋅bx=-10-1±1+20(6)Multiplyx=-10-1±1+120Add termsx=-10-1±121Calculate root x=-10-1±11 This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.x=-10-1±11 x1=-10-1+11x2=-10-1−11 x1=-1010x2=-10-12 x1=-1x2=56Therefore, we obtained two solutions to the given system of equations (-1,-7) and (56,-7). Checking Our Answer info Checking the answer We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (-1,-7). We will substitute -1 and -7 for x and y, respectively, in Equation (I) and Equation (II). {y=-5x2+x−1y=-7(I), (II): x=-1, y=-7{-7=-5(-1)2+(-1)−1-7=-7 Simplify (I): (-a)2=a2{-7=-5(1)2+(-1)−1-7=-7(I): 1a=1{-7=-5(1)+(-1)−1-7=-7(I): a⋅1=a{-7=-5+(-1)−1-7=-7(I): a+(-b)=a−b{-7=-5−1−1-7=-7(I): Subtract terms {-7=-7 ✓-7=-7 ✓ Since both equations produce true statements, the solution (-1,-7) is correct. Let's now check (56,-7). {y=-5x2+x−1y=-7(I), (II): x=56, y=-7{-7=-5(56)2+(56)−1-7=-7 Simplify (I): Calculate power{-7=-5(2536)+56−1-7=-7(I): a⋅cb=ca⋅b{-7=-255(36)+56−1-7=-7(I): ba=b/5a/5{-7=-536+56−1-7=-7(I): Add fractions{-7=-530−1-7=-7(I): Calculate quotient{-7=-6−1-7=-7(I): Subtract term {-7=-7 ✓-7=-7 ✓ Since again both equations produce true statements, the solution (56,-7) is also correct. | |

Exercises 11 To solve this system of equations, we will graph the given functions. ⎩⎪⎨⎪⎧y=214x+1y=x2−2x+4 Let's simplify the first equation. y=214x+1a1=a-1y=2-14x+1Write as a powery=2-1(22)x+1(am)n=am⋅ny=2-122x+1am⋅an=am+ny=22x−1+1 Therefore, we can rewrite our system of equations. {y=22x−1+1y=x2−2x+4(I)(II) Now, we will make a table of values for both equations.Equation (I)Equation (II) x22x−1+1y=22x−1+1xx2−2x+4y=x2−2x+4 -222(-2)−1+11.03-2(-2)2−2(-2)+412 -122(-1)−1+11.12-1(-1)2−2(-1)+47 022(0)−1+11.50(0)2−2(0)+44 122(1)−1+131(1)2−2(1)+43 222(2)−1+192(2)2−2(2)+44 Next, we will plot the points and connect them with a smooth curve for Equation (I) and Equation (II).We can see from both the table of values and the graph that there is exactly one point of intersection (1,3). Finally, we found the solution to the system. {x=1y=3 | |

Exercises 12 For this exercise, it is important to look through the point criteria and solve for each one, then add the points together.Maximum height can be found by using the vertex formula. Time in air can be found by solving for t when h=0. Perfect landing is stated and we can assign 25 points.Maximum Height Let's start by finding the vertex's t-value of the given equation, h=-16t2+28t+8. We can use the formula for the axis of symmetry. t=2a-ba=-16, b=28t=2(-16)-28Calculate quotient and productt=0.875 Now we can use that value for x to find the vertex's h-value for this equation. h=-16t2+28t+8t=0.875h=-16(0.875)2+28(0.875)+8Calculate power and producth=-12.25+24.5+8Add termsh=20.25 From this we can say that the skier reached a maximum height of 20.25 feet and, therefore, earned and additional 20.25 points as they score 1 point per foot.Time in the Air To find the amount of time the skier spent in the air, we need to calculate how long it took to land. This means we need to find the value of t when h=0. Let's use the Quadratic Formula to do that. 0=-16t2+28t+8Use the Quadratic Formula: a=-16,b=-28,c=8t=2(-16)-(26)±(26)2−4(-16)(8) Simplify right-hand side Calculate power and productt=-32-26±676−(-512)Subtract termt=-32-26±1188Calculate root t≈-32-26±34.48 The solutions for this equation are t≈-32-26±34.48. Let's separate them into the positive and negative cases.t≈-32-26±34.48. t1≈t2≈ -32-26+34.48-32-26−34.48 -328.47-32-60.47 -0.261.88 Using the Quadratic Formula, we found that the solutions are t1=-0.26 and t2=1.88. We can ignore the negative solution and say that the skier was in the air for 1.88 sec. Since the skier gets 5 points per second, we can multiply 1.88⋅5 and assign 9.4 points for time in the air.Perfect Landing The exercise states that the skier had a perfect landing which rewards another 25 points. Let's add all the categories together and find the number of points the skier can earn. Max Height↓20.25++Time in Air↓9.4++Perfect Landing↓25==Total Points↓54.65 | |

Exercises 13 For this exercise, we need to use the function provided to determine how long the riders experience free fall. Let's use the time the brakes are activated to represent the end of free fall. We will substitute h=105, into the equation and solve for t. h=-16t2+265h=105105=-16t2+265 Solve for t LHS−265=RHS−265-160=-16t2LHS/(-16)=RHS/(-16)10=t2Rearrange equationt2=10LHS=RHS t=±10 In this exercise, free fall starts when the riders are dropped from 265 ft at t=0. Furthermore, t represents the time after the ride starts, so we can ignore the negative solution. Since 10≈3.16, we can say that the riders experience free fall for 3.16 sec. | |

Exercises 14 In this exercise, we are shown a painting with expressions for length and width and asked to find and expression for the area. Let's use the rectangle area formula and substitute ℓ=30x7 in and w=336 in. A=ℓ⋅wℓ=30x7, w=336A=30x7⋅336 Simplify right-hand side Multiply fractionsA=33630x7ba=baA=3610x7Simplify radical and terms A=36x310x Since this expression cannot be simplified further, we can say a rectangle with sides 30x7 in and 336 in has an area of 36x310x in2. | |

Exercises 15 The x-intercepts of a graph are the zeros of the function. Therefore, the number of times the graph intersects the x-axis equals the number of x-values for which y(x)=0. Let's take a look at the given function. y=5x2+10x+5 By setting the function equal to zero we obtain a quadratic equation we can solve for the zeros of the function. y(x)=0⇒5x2−10x+5=0 Recall that we can know how many solutions a quadratic equation has by evaluating 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. Let's give it a try. We can identify a, b, and c by comparing our equation to the standard form of a quadratic equation. ax2+bx+c=05x2+(-10)x+5=0 As we can see, a=5, b=-10, and c=5. Let's now evaluate the discriminant. b2−4acSubstitute values(-10)2−4(5)(5) Simplify Calculate power100−4(5(5)Multiply100−100Subtract term 0 Since the discriminant is zero, the equation has 1 real solution and our function has just one zero. Consequently, its graph intersects the x-axis just one time. Alternative solution info Finding the zeros by factoring To know how many zeros a function has, it useful to rewrite it in its factored form. Notice that our function has a Greatest Common Factor of 5. Then, we can start by factoring 5 out. y=5x2+10x+5Factor out 5y=5(x2−2x+1)a2−2ab+b2=(a−b)2y=5(x−1)2 Now that we know its factored form, by setting the function equal to zero we can identify the function's zeros right away. y(x)=0⇒5(x−1)2=0 As we can see, the function becomes zero only when x=1. Therefore, it has only one zero, and consequently, its graph intersects the x-axis one time. | |

Exercises 16 | |

Exercises 17 |

##### Other subchapters in Solving Quadratic Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Properties of Radicals
- Solving Quadratic Equations by Graphing
- Solving Quadratic Equations Using Square Roots
- Quiz
- Solving Quadratic Equations by Completing the Square
- Solving Quadratic Equations Using the Quadratic Formula
- Solving Nonlinear Systems of Equations
- Chapter Review
- Cumulative Assessment