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###### Exercises
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Exercises 1 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 72p7​ Split into factors Rewrite 72 as 9⋅89⋅8⋅p7​Rewrite 8 as 4⋅29⋅4⋅2⋅p7​Rewrite p7 as p6⋅p9⋅4⋅2⋅p6⋅p​Rewrite p6 as p2⋅p2⋅p2 9⋅4⋅2⋅p2⋅p2⋅p2⋅p​a⋅b​=a​⋅b​9​⋅4​⋅2​⋅p2​⋅p2​⋅p2​⋅p​Calculate root3⋅2⋅2​⋅p2​⋅p2​⋅p2​⋅p​a2​=a3⋅2⋅2​⋅p⋅p⋅p⋅p​ Multiply a⋅a⋅a=a33⋅2⋅2​⋅p3⋅p​Multiply6p3⋅2​⋅p​a​⋅b​=a⋅b​ 6p32p​ Remember that we assume that the variable p is positive.
Exercises 2 Let's start simplifying by using the quotient property of radicals. 7y45​​=7y​45​​​ To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by itself. 7y​45​​ba​=b⋅7y​a⋅7y​​7y​⋅7y​45​⋅7y​​a​⋅a​=a7y45​⋅7y​​ We know that we have successfully rationalized the denominator because the radical has been eliminated. However, our fraction can still be simplified a bit further. Let's try! 7y45​⋅7y​​Split into factors7y9⋅5​⋅7y​​a⋅b​=a​⋅b​7y9​⋅5​⋅7y​​Calculate root7y3⋅5​⋅7y​​a​⋅b​=a⋅b​7y3⋅5⋅7y​​Multiply7y335y​​
Exercises 3 We will start simplifying this expression by expanding the fraction so that the denominator is a perfect cube. To do so, we have to multiply both the numerator and the denominator by 2. 34125x11​​ba​=b⋅2a⋅2​34⋅2125x11⋅2​​Multiply38250x11​​3ba​​=3b​3a​​38​3250x11​​Calculate root23250x11​​ Next we will take care of the numerator. We will rewrite the number under the cubic root as the product of at least one perfect cube factor. 23250x11​​ Split into factors Split into factors23125⋅2x11​​Write as a power2353⋅2x11​​Write as a sum2353⋅2x9+2​​am+n=am⋅an 2353⋅2⋅x9⋅x2​​3a⋅b​=3a​⋅3b​2353​⋅32​⋅3x9​⋅3x2​​nan​=a{n}25⋅32​⋅3x9​⋅3x2​​nam​=anm​25⋅32​⋅x3⋅3x2​​ Multiply Commutative Property of Multiplication25⋅x3⋅32​⋅3x2​​3a​⋅3b​=3a⋅b​25⋅x3⋅32⋅x2​​Multiply 25x332x2​​ Remember that we assume that the variable x is positive.
Exercises 4 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a+ba−b a−ba+bIn this case, the conjugate of the denominator is 6​−2. 6​+28​ba​=b⋅(6​−2)a⋅(6​−2)​(6​+2)(6​−2)8(6​−2)​(a+b)(a−b)=a2−b2(6​)2−228(6​−2)​(a​)2=a6−228(6​−2)​Calculate power6−48(6​−2)​Subtract term28(6​−2)​ba​=b/2a/2​4(6​−2)Distribute 446​−8
Exercises 6 To simplify the given radical expression, we first need to rewrite the radicands so that they have exponents that match the index of the radicals. Then we will consider the properties for combining radical expressions when they are part of a sum or difference. anx​+bnx​=(a+b)nx​anx​−bnx​=(a−b)nx​​ Notice that radicals can only be added or subtracted when the index and the value inside are exactly the same. Let's simplify our radicals to see if we can create like terms. 1532​−2354​Split into factors1532​−2327⋅2​3a⋅b​=3a​⋅3b​1532​−2327​⋅32​Calculate root1532​−2⋅3⋅32​Multiply1532​−632​ We did it! Now that the terms have the same radical expression, we can subtract them. 1532​−632​Factor out 32​(15−6)32​Subtract term932​ The simplest form of the radical expression is 932​.
Exercises 7 We will use the square of a binomial pattern. (a+b)2=a2+2ab+b2​ Let's determine the values of a and b in our case. (37​+5)2=(37​)2+2(37​)(5)+52​ Finally we will simplify the obtained expression. (37​)2+2(37​)(5)+52 Calculate power (a⋅b)m=am⋅bm32⋅(7​)2+2(37​)(5)+52(a​)2=a32⋅7+2(37​)(5)+52Calculate power 9⋅7+2(37​)(5)+25Multiply63+307​+25Commutative Property of Addition63+25+307​Add terms88+307​
Exercises 10 We are asked to solve the given quadratic equation by graphing. There are three steps to solving a quadratic equation by graphing.Write the equation in standard form, ax2+bx+c=0. Graph the related function y=ax2+bx+c. Find the x-intercepts, if any.The solutions of ax2+bx+c=0 are the x-intercepts of the graph of y=ax2+bx+c. Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation. x2−2x=-4⇔x2−2x+4=0​ Now we can identify the function related to the equation. Equation: Related Function: ​x2−2x+4=0y=x2−2x+4​Graphing the Related Function To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=x2−2x+4⇔y=1x2+(-2)x+4​ We can see that a=1, b=-2, and c=4. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab​. Since we already know the values of a and b, we can substitute them into the formula. x=-2ab​a=1, b=-2x=-2(1)-2​ Simplify right-hand side a⋅1=ax=-2-2​-b-a​=ba​x=22​aa​=1 x=1 The axis of symmetry of the parabola is the vertical line with equation x=1.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab​,f(-2ab​))​ Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=1. Thus, the x-coordinate of the vertex is also 1. To find the y-coordinate, we need to substitute 1 for x in the given equation. y=x2−2x+4x=1y=12−2(1)+4 Simplify right-hand side 1a=1y=1−2(1)+4a⋅1=ay=1−2+4Add and subtract terms y=3 We found the y-coordinate, and now we know that the vertex is (1,3).Identifying the y-intercept and its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,4). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.Finding the x-intercepts Let's identify the x-intercepts of the graph of the related function, if there are any.We can see that the parabola does not intercept the x-axis. Therefore, the equation x2−2x=-4 has no solutions.
Exercises 11 We are asked to solve the given quadratic equation by graphing. There are three steps to solving a quadratic equation by graphing.Write the equation in standard form, ax2+bx+c=0. Graph the related function y=ax2+bx+c. Find the x-intercepts, if any. The solutions of ax2+bx+c=0 are the x-intercepts of the graph of y=ax2+bx+c. Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation. -8x−16=x2LHS−x2=RHS−x2-8x−16−x2=0Commutative Property of Addition-x2−8x−16=0 Now we can identify the function related to the equation. Equation: Related Function: ​-x2−8x−16=0y=-x2−8x−16​Graphing the Related Function To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=-x2−8x−16⇔y=-1x2+(-8)x+(-16)​ We can see that a=-1, b=-8, and c=-16. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab​. Since we already know the values of a and b, we can substitute them into the formula. x=-2ab​a=-1, b=-8x=-2(-1)-8​ Simplify right-hand side a(-b)=-a⋅bx=--2-8​-b-a​=ba​x=-28​Calculate quotient x=-4 The axis of symmetry of the parabola is the vertical line with equation x=-4.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab​,f(-2ab​))​ Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=-4. Thus, the x-coordinate of the vertex is also -4. To find the y-coordinate, we need to substitute -4 for x in the given equation. y=-x2−8x−16x=-4y=-(-4)2−8(-4)−16 Simplify right-hand side (-a)2=a2y=-16−8(-4)−16-a(-b)=a⋅by=-16+32−16Add and subtract terms y=0 We found the y-coordinate, and now we know that the vertex is (-4,0).Identifying the y-intercept and its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,-16). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=-1, which is negative, the parabola will open downwards. Let's connect the three points with a smooth curve.Finding the x-intercepts Let's identify the x-intercepts of the graph of the related function.We can see that the parabola intercepts the x-axis once. The point of intersection is (-4,0), it is also the vertex of the parabola. The equation -8x−16=x2 has one solution, x=-4. Checking Our Answer info Checking Our Solutions We can check our solution by substituting the value into the given equation. -8x−16=x2x=-4-8(-4)−16=?(-4)2 Simplify -a(-b)=a⋅b32−16=?(-4)2Subtract term16=?(-4)2(-a)2=a2 16=16 Since 16=16, we know that x=-4 is a solution of the equation.
Exercises 12 We want to use the given graph to determine the zeros of the function f. The zeros are the x-coordinates of the points of intersection of the graph and the x-axis. Let's consider the graph.This graph crosses the x-axis three times, so it has three zeros. We can see that those zeros are x=-3, x=-1, and x=1.
Exercises 13 To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of a, b, and c. y=x2+2x−5⇔y=1x2+2x+(-5)​ We can see that a=1, b=2, and c=-5. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab​. Since we already know the values of a and b, we can substitute them into the formula. x=-2ab​a=1, b=2x=-2(1)2​ Simplify right-hand side a⋅1=ax=-22​aa​=1 x=-1 The axis of symmetry of the parabola is the vertical line with equation x=-1.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab​,f(-2ab​))​ Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=-1. Thus, the x-coordinate of the vertex is also -1. To find the y-coordinate, we need to substitute -1 for x in the given equation. y=x2+2x−5x=-1y=(-1)2+2(-1)−5 Simplify right-hand side (-a)2=a2y=1+2(-1)−5a(-b)=-a⋅by=1−2−5Subtract terms y=-6 We found the y-coordinate, and now we know that the vertex is (-1,-6).Identifying the y-intercept and its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,-5). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.By looking at the graph, we can state approximated values for the zeros of the function. We can see that the parabola intercepts the x-axis at x=-3.4 and x=1.4.
Exercises 14 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. x2+5=17LHS−5=RHS−5x2=12LHS​=RHS​x=±12​Use a calculatorx=±3.46410…x≈±3.46 We found that x≈±3.46. Thus, there are two solutions for the equation, x≈3.46 and x≈-3.46.
Exercises 15 Let's start by isolating x2 on the left-hand side of the equation. x2−14=-14LHS+14=RHS+14x2−14+14=-14+14Add termsx2=0LHS​=RHS​x=0
Exercises 16 To solve the given equation by taking the square roots, we need to consider the positive and negative solutions. (x+2)2=64LHS​=RHS​x+2=±64​Calculate rootx+2=±8LHS−2=RHS−2x=±8−2State solutionsx=8−2x=-8−2​(I)(II)​(I), (II):  Subtract termsx=6x=-10​ There are two solutions for the equation, x=6 and x=-10. 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=6. (x+2)2=64x=6(6+2)2=?64 Simplify Add terms(8)2=?64Calculate power 64=64 ✓ Since we obtained a true statement, x=6 is a solution to the equation. Let's check if x=-10 is also a solution. (x+2)2=64x=-10(-10+2)2=?64 Simplify Add terms(-8)2=?64(-a)2=a282=?64Calculate power 64=64 ✓ Since we obtained a true statement, x=-10 is a solution to the equation.
Exercises 17 Let's start by isolating x2 on the left-hand side of the equation. 4x2+25=-75LHS−25=RHS−254x2=-100LHS/4=RHS/4x2=-25 The square of a real number cannot be negative. Therefore, the equation has no real solutions.
Exercises 18 We will solve the given equation by taking the square roots. (x−1)2=0LHS​=RHS​x−1=±0Zero Property of Multiplicationx−1=0LHS+1=RHS+1x=1 The only solution to the given equation is x=1. Checking Our Answer info Checking our answer We can check our answer by substituting it for x in the given equation. (x−1)2=0x=1(1−1)2=?0 Simplify Subtract term02=?0Calculate power 0=0 ✓ Since we obtained a true statement, we know that x=1 is a solution of the equation.
Exercises 19 To solve the given equation by taking square roots, we need to consider the positive and negative solutions. 19=30−5x2Rearrange equation30−5x2=19LHS−30=RHS−30-5x2=-11LHS/(-5)=RHS/(-5)x2=-5-11​-b-a​=ba​x2=511​LHS​=RHS​x=±511​​Use a calculatorx=±1.48323…x≈±1.48 We found that x≈±1.48. Thus, there are two solutions for the equation, x≈1.48 and x≈-1.48.
Exercises 20 We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation. x2+6x−40=0​ In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=6. Let's now calculate (2b​)2. (2b​)2b=6(26​)2 Simplify Calculate quotient32Calculate power 9 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−40=0 Factor LHS+9=RHS+9x2+6x+9−40=9Split into factorsx2+2x(3)+9−40=9Write as a powerx2+2x(3)+32−40=9a2+2ab+b2=(a+b)2 (x+3)2−40=9LHS+40=RHS+40(x+3)2=49LHS​=RHS​x+3=±49​Calculate rootx+3=±7LHS−3=RHS−3x=-3±7State solutionsx=-3+7x=-3−7​(I)(II)​(I), (II):  Add and subtract termsx=4x=-10​ Both x=4 and x=-10 are solutions to the equation.
Exercises 21 We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation. x2+2x+5=4​ 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=2(22​)2 Simplify aa​=1121a=1 1 Next, we will add (2b​)2=1 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+2x+5=4 Factor LHS+1=RHS+1x2+2x+1+5=4+1Split into factorsx2+2x(1)+1+5=4+1Write as a powerx2+2x(1)+12+5=4+1a2+2ab+b2=(a+b)2 (x+1)2+5=4+1LHS−5=RHS−5(x+1)2=4+1−5Add and subtract terms(x+1)2=0LHS​=RHS​x+1=0LHS−1=RHS−1x=-1
Exercises 22 We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation. 2x2−4x=10​ Now let's divide each side by 2 so the coefficient of x2 will be 1. 2x2−4x=10LHS/2=RHS/222x2−4x​=210​ Simplify Write as a difference of fractions22x2​−24x​=210​ca⋅b​=ca​⋅b22​x2−24​x=210​Calculate quotient x2−2x=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=-2(2-2​)2 Simplify Put minus sign in front of fraction(-22​)2aa​=1(-1)2(-a)2=a2 1 Next, we will add (2b​)2=1 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. x2−2x=5 Factor LHS+1=RHS+1x2−2x+1=5+1Split into factorsx2−2x(1)+1=5+1Write as a powerx2−2x(1)+12=5+1a2−2ab+b2=(a−b)2 (x−1)2=5+1Add terms(x−1)2=6LHS​=RHS​x−1=±6​LHS+1=RHS+1x=1±6​Use a calculatorx=1±2.449…x≈1±2.45State solutionsx≈1+2.45x≈1−2.45​(I)(II)​(I), (II):  Add and subtract termsx≈3.45x≈-1.45​
Exercises 23 To find out whether the function has a minimum or maximum value, we need to look at the value of a. y=ax2+bx+c⇒y=-1x2+6x+(-1)​ Because a=-1 is negative, we know that the parabola opens down and the y-coordinate of the vertex is the maximum value. We can find it by rewriting the given quadratic function in vertex form. We will start by factoring out a so the coefficient of x2 will be 1. y=-x2+6x−1⇕y=-(x2−6x+1)​ We can see that b=-6. Let's use this to calculate (2b​)2. (2b​)2b=-6(2-6​)2 Simplify Put minus sign in front of fraction(-26​)2Calculate quotient(-3)2(-a)2=a2 32 Let's now factor the function by completing the square. We will first add (2b​)2=32. Note that we also have to subtract 32 to leave the function unchanged. If you need further explanations on how to factor this kind of expressions, please refer to this exercise or this exercise. y=-(x2−6x+1)a=a+32−32y=-(x2−6x+32+1−32)a2−2ab+b2=(a−b)2y=-((x−3)2+1−32)Calculate powery=-((x−3)2+1−9)Subtract termy=-((x−3)2−8)Distribute -1y=-(x−3)2+8 Now we know that the vertex form is y=-(x−3)2+8 and the vertex of the parabola is (3,8). Therefore, the maximum value of the function is 8.
Exercises 24 To find out whether the function has a minimum or maximum value, we need to look at the value of a. f(x)=ax2+bx+c⇒f(x)=1x2+4x+11​ Because a=1 is positive, we know that the parabola opens up and the y-coordinate of the vertex is the minimum value. We can find it by rewriting the given quadratic function in vertex form. We can see that b=4. Let's use this to calculate (2b​)2. (2b​)2b=4(24​)2Calculate quotient22 Let's now factor the function by completing the square. We will first add (2b​)2=22. Note that we also have to subtract 22 to leave the function unchanged. If you need further explanations on how to factor this kind of expressions, please refer to this exercise or this exercise. f(x)=x2+4x+11a=a+22−22f(x)=x2+4x+22+11−22a2+2ab+b2=(a+b)2f(x)=(x+2)2+11−22Calculate powerf(x)=(x+2)2+11−4Subtract termf(x)=(x+2)2+7 Now we know that the vertex form is f(x)=(x−(-2))2+7 and the vertex of the parabola is (-2,7). Therefore, the minimum value of the function is 7.
Exercises 25 To find out whether the function has a minimum or maximum value, we need to look at the value of a. y=ax2+bx+c⇒y=3x2+(-24)x+15​ Because a=3 is positive, we know that the parabola opens up and the y-coordinate of the vertex is the minimum value. We can find it by rewriting the given quadratic function in vertex form. We will start by factoring out a so the coefficient of x2 will be 1. y=3x2−24x+15⇕y=3(x2−8x+5)​ We can see that b=-8. Let's use this to calculate (2b​)2. (2b​)2b=-8(2-8​)2Calculate quotient(-4)2(-a)2=a242 Let's now factor the function by completing the square. We will first add (2b​)2=42. Note that we also have to subtract 42 to leave the function unchanged. If you need further explanations on how to factor this kind of expressions, please refer to this exercise or this exercise. y=3(x2−8x+5)a=a+42−42y=3(x2−8x+42+5−42)a2−2ab+b2=(a−b)2y=3((x−4)2+5−42)Calculate powery=3((x−4)2+5−16)Subtract termy=3((x−4)2−11)Distribute 3y=3(x−4)2−33 Now we know that the vertex form is y=3(x−4)2+(-33) and the vertex of the parabola is (4,-33). Therefore, the minimum value of the function is -33.
Exercises 26 To find the perimeter of the credit card we need to know its width and its length. Let's determine these values using the given information. Recall the formula for the area of a rectangle. A=ℓw​ Here ℓ is the length of the rectangle and w is the width of the rectangle. In our case A=46.75. We also know that the width w is 3 centimeters shorter than the length ℓ, so w=ℓ−3. We will substitute these values into the formula. By doing so we obtain an equation that can be solved for ℓ. A=ℓwA=46.75, w=ℓ−346.75=ℓ(ℓ−3)Rearrange equationℓ(ℓ−3)=46.75Distribute ℓℓ2−3ℓ=46.75 Let's solve this equation by completing the square. We will follow five steps.Find one-half of b, the coefficient of ℓ. Square the result from Step 1. Add the result from Step 2 to both sides of the equation. Factor the left-hand side of the equation from Step 3 as the square of a binomial, ℓ2+bℓ+(2b​)2=(ℓ+2b​)2. Take a square root of both sides of the obtained equation. Let's get started! We will identify the value of b first. ℓ2−3ℓ=46.75⇔ℓ2+(-3)ℓ=46.75​ Therefore, b=-3. Next we will find one half of b. 2-3​=-1.5​ Moving on to Step 2, we have to square our result. (-1.5)2=1.52=2.25​ Let's add 2.25 to both sides of the equation. ℓ2−3ℓ=46.75LHS+2.25=RHS+2.25ℓ2−3ℓ+2.25=49 Now we are able to write the left-hand side of the equation as the square of a binomial. ℓ2−3ℓ+2.25=49⇔(ℓ−1.5)2=49​ Finally, we will take a square root of both sides of the equation. Remember to calculate the positive and negative square roots. (ℓ−1.5)2=49LHS​=RHS​ℓ−1.5=±49​Calculate rootℓ−1.5=±7LHS+1.5=RHS+1.5ℓ=1.5±7 The solutions to the equation are 1.5+7=8.5 and 1.5−7=-5.5. Since ℓ represents the length of the card, it cannot be negative. Therefore, ℓ=8.5 centimeters and w=8.5−3=5.5 centimeters. Now we can calculate the perimeter of the card. We will use the formula for the perimeter of a rectangle. P=2ℓ+2w​ Let's substitute ℓ=8.5 and w=5.5! P=2ℓ+2wℓ=8.5, w=5.5P=2(8.5)+2(5.5)MultiplyP=17+11Add termsP=28 The perimeter of the credit card is 28 centimeters.
Exercises 27 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+2x−15=0⇔1x2+2x+(-15)=0​ We see that a=1, b=2, and c=-15. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-2±22−4(1)(-15)​​ Solve for x and Simplify Calculate powerx=2(1)-2±4−4(1)(-15)​​Multiplyx=2-2±4−4(-15)​​-a(-b)=a⋅bx=2-2±4+60​​Add termsx=2-2±64​​Calculate root x=2-2±8​ The solutions for this equation are x=2-2±8​. Let's separate them into the positive and negative cases.x=2-2±8​ x1​=2-2+8​x2​=2-2−8​ x1​=26​x2​=2-10​ x1​=3x2​=-5 Using the Quadratic Formula, we found that the solutions of the given equation are x1​=3 and x2​=-5.
Exercises 28 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−x+8=16LHS−16=RHS−162x2−x−8=0 Now, we can identify the values of a, b, and c. 2x2−x−8=0⇔2x2+(-1)x+(-8)=0​ We see that a=2, b=-1, and c=-8. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(2)-(-1)±(-1)2−4(2)(-8)​​ Solve for x and Simplify -(-a)=ax=2(2)1±(-1)2−4(2)(-8)​​Calculate powerx=2(2)1±1−4(2)(-8)​​Multiplyx=41±1−8(-8)​​-a(-b)=a⋅bx=41±1+64​​Add termsx=41±65​​Use a calculatorx=41±8.062…​Round to nearest integer x≈41±8​ The solutions for this equation are x≈41±8​. Let's separate them into the positive and negative cases.x≈41±8​ x1​≈41+8​x2​≈41−8​ x1​≈49​x2​≈4-7​ x1​≈2.25x2​≈-1.75 x1​≈2.3x2​≈-1.8 Using the Quadratic Formula, we found that the solutions of the given equation are x1​≈2.3 and x2​≈-1.8.
Exercises 29 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+10x=5LHS−5=RHS−5-5x2+10x−5=0Factor out -5-5(x2−2x+1)=0LHS/(-5)=RHS/(-5)x2−2x+1=0 Now, we can identify the values of a, b, and c. x2−2x+1=0⇔1x2+(-2)x+1=0​ We see that a=1, b=-2, and c=1. Let's substitute these values into the Quadratic Formula. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-2)±(-2)2−4(1)(1)​​ Solve for x and Simplify -(-a)=ax=2(1)2±(-2)2−4(1)(1)​​Calculate powerx=2(1)2±4−4(1)(1)​​Multiplyx=22±4−4​​Subtract termx=22±0​​Calculate root x=22±0​ Since adding or subtracting zero doesn't change the value of a number, the numerator will simplify to 2. Therefore, we will get only one value of x. x=22​⇔x=1​ Using the Quadratic Formula, we found that the solution of the given equation is x=1.
Exercises 30 To determine the number of x-intercepts of the graph of the given quadratic function, we will use the discriminant of the quadratic equation to which the function is related. Function: Related Equation: ​y=-x2+6x−9-x2+6x−9=0​ 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. Let's identify the values of a, b, and c. -x2+6x−9=0⇔-1x2+6x+(-9)=0​ Next, let's evaluate the discriminant. b2−4acSubstitute values62−4(-1)(-9) Simplify Calculate power36−4(-1)(-9)(-a)(-b)=a⋅b36−4(9)Multiply36−36Subtract term 0 Since the discriminant is 0, the graph of the quadratic function has one x-intercept. Extra info Further information If the discriminant is greater than zero, the graph of the related function will have two x-intercepts. If it is equal to zero, the graph will have one x-intercept. Finally, if the discriminant is less than zero, the graph will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21690_1999490017_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,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, 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,"\\text{One }x\\text{-intercept}",{ mathMode:true, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{One }x\\text{-intercept}",{ mathMode:true, 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,"\\text{No }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{No }x\\text{-intercepts}",{ mathMode:true, 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: "Solution21690_1999490017_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21690_1999490017_l", "Solution21690_1999490017_p", 1, code); }); } ); } window.JXQtable["Solution21690_1999490017_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn819224116_1873863618').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 ($) {$('#jsxbtn1852816663_1451832406').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 ($) {$('#jsxbtn589511877_1784878957').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 ($) {$('#jsxbtn209247469_1957677666').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 31 To determine the number of x-intercepts of the graph of the given quadratic function, we will use the discriminant of the quadratic equation to which the function is related. Function: Related Equation: ​2x2+4x+8=0y=2x2+4x+8​ 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. Let's identify the values of a, b, and c. 2x2+4x+8=0​ Next, let's evaluate the discriminant. b2−4acSubstitute values42−4(2)(8) Simplify Calculate power16−4(2)(8)Multiply16−64Subtract term -48 Since the discriminant is -48, the graph of the quadratic function has no x-intercepts. Extra info Further information If the discriminant is greater than zero, the graph of the related function will have two x-intercepts. If it is equal to zero, the graph will have one x-intercept. Finally, if the discriminant is less than zero, the graph will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21691_839953621_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,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, 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,"\\text{One }x\\text{-intercept}",{ mathMode:true, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{One }x\\text{-intercept}",{ mathMode:true, 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,"\\text{No }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{No }x\\text{-intercepts}",{ mathMode:true, 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: "Solution21691_839953621_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21691_839953621_l", "Solution21691_839953621_p", 1, code); }); } ); } window.JXQtable["Solution21691_839953621_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn387854403_15947615').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 ($) {$('#jsxbtn198674819_196837693').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 ($) {$('#jsxbtn538906341_2055672112').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 ($) {$('#jsxbtn1535386124_716966460').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 32 To determine the number of x-intercepts of the graph of the given quadratic function, we will use the discriminant of the quadratic equation to which the function is related. Function: Related Equation: ​-21​x2+2x=0y=-21​x2+2x​ 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. Let's identify the values of a, b, and c. -21​x2+2x=0⇔-21​x2+2x+0=0​ Next, let's evaluate the discriminant. b2−4acSubstitute values22−4(-21​)(0) Simplify Calculate power4−4(-21​)(0)Zero Property of Multiplication4−0Subtract term 4 Since the discriminant is 4, the graph of the quadratic function has two x-intercepts. Extra info Further information If the discriminant is greater than zero, the graph of the related function will have two x-intercepts. If it is equal to zero, the graph will have one x-intercept. Finally, if the discriminant is less than zero, the graph will have no x-intercepts. window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21692_248583051_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,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#628aff", fontsize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"\\text{Two }x\\text{-intercepts}",{ mathMode:true, 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,"\\text{One }x\\text{-intercept}",{ mathMode:true, flag:true, flagColor:"#00c997", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{One }x\\text{-intercept}",{ mathMode:true, 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,"\\text{No }x\\text{-intercepts}",{ mathMode:true, flag:true, flagColor:"#ff8a62", fontsize:0.8 } ); el.txt2 = b2.txt(0,0.9,"\\text{No }x\\text{-intercepts}",{ mathMode:true, 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: "Solution21692_248583051_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21692_248583051_l", "Solution21692_248583051_p", 1, code); }); } ); } window.JXQtable["Solution21692_248583051_l"] = true;b2−4ac>0 window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn226073767_271714685').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 ($) {$('#jsxbtn35864904_1036114578').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 ($) {$('#jsxbtn1335145426_1178064067').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 ($) {$('#jsxbtn2079343797_1091957924').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );
Exercises 33 We want to solve the given system of equations using the Substitution Method. {y=x2−2x−4y=-5​(I)(II)​ Note that Equation (II) is already solved for the variable y. This allows us to substitute y=-5 in Equation (I). {y=x2−2x−4y=-5​(I):  y=-5{-5=x2−2x−4y=-5​(I):  LHS+5=RHS+5{0=x2−2x+1y=-5​(I):  Rearrange equation{x2−2x+1=0y=-5​ Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x2−2x+1=0⇔1x2+(-2)x+1=0​ Now, recall the Quadratic Formula. x=2a-b±b2−4ac​​​ We can substitute a=1, b=-2, and c=1 into this formula to solve the quadratic equation. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-2)±(-2)2−4(1)(1)​​ Solve for x -(-a)=ax=2(1)2±(-2)2−4(1)(1)​​Calculate powerx=2(1)2±4−4(1)(1)​​Multiplyx=22±4−4​​Subtract termx=22±0​​Calculate root x=22±0​ Since adding or subtracting zero doesn't change the value of a number, the numerator will simplify to 2. Therefore, we will get only one value of x. x=22​⇔x=1​ We found that x=1 and y=-5. Therefore, the only solution, which is the point of intersection of the graphs of the two functions, is (1,-5).Checking Our Answer info Checking the answer We can check our answer by substituting the point into both equations. If they produce true statements, our solutions are correct. We will substitute 1 and -5 for x and y, respectively, in Equation (I) and Equation (II). {y=x2−2x−4y=-5​(I), (II):  x=1, y=-5{-5=?12−2(1)−4-5=-5​ Simplify (I):  1a=1{-5=?1−2(1)−4-5=-5​(I):  Multiply{-5=?1−2−4-5=-5​(I):  Subtract terms {-5=-5 ✓-5=-5 ✓​ Since both equations produced true statements, the solution (1,-5) is correct.
Exercises 34 We want to solve the given system of equations using the Substitution Method. {y=x2−9y=2x+5​(I)(II)​ The y-variable is isolated in Equation (II). This allows us to substitute its value, 2x+5, for y in Equation (I). {y=x2−9y=2x+5​(I):  y=2x+5{2x+5=x2−9y=2x+5​ (I):  Simplify (I):  LHS−5=RHS−5{2x=x2−14y=2x+5​(I):  LHS−2x=RHS−2x{0=x2−2x−14y=2x+5​(I):  Rearrange equation {x2−2x−14=0y=2x+5​ Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x2−2x−14=0⇔1x2+(-2)x+(-14)=0​ Now, recall the Quadratic Formula. x=2a-b±b2−4ac​​​ We can substitute a=1, b=-2, and c=-14 into this formula to solve the quadratic equation. x=2a-b±b2−4ac​​Substitute valuesx=2(1)-(-2)±(-2)2−4(1)(-14)​​ Solve for x -(-a)=ax=2(1)2±(-2)2−4(1)(-14)​​Calculate powerx=2(1)2±4−4(1)(-14)​​a⋅1=ax=22±4−4(-14)​​-a(-b)=a⋅bx=22±4+56​​Add termsx=22±60​​Split into factorsx=22±4⋅15​​a⋅b​=a​⋅b​x=22±4​15​​Calculate rootx=22±215​​Factor out 2x=22(1±15​)​Cancel out common factors x=1±15​ 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. Therefore, we have that x1​=1+15​ and x2​=1−15​. Now, consider Equation (II). y=2x+5​ We can substitute x1​=1+15​ and x2​=1−15​ in the above equation to find the values for y. Let's start with x1​=1+15​. y=2x+5x=1+15​y=2(1+15​)+5Distribute 2y=2+215​+5Add termsy=7+215​ We found that y=7+215​ when x=1+15​. One solution to the system is (1+15​,7+215​). To find the other solution, we will substitute 1−15​ for x in Equation (II) again. y=2x+5x=1−15​y=2(1−15​)+5Distribute 2y=2−215​+5Add termsy=7−215​ We found that y=7−215​ when x=1−15​. Therefore, our second solution is (1−15​,7−215​). Finally, if desired, we can approximate our solutions to two decimal places. (1+15​,7+215​)(1−15​,7−215​)​≈(4.87,14.75)≈(-2.87,-0.75)​
Exercises 35 The solutions to the system of equations will be the points of intersection of the graphs of the functions. {y=2(21​)x−5y=-x2−x+4​ To find these points, we will graph the functions using a graphing calculator. To do that we should enter them in the calculator by pushing Y=​ and writing their rules in the first two rows.Next, by pushing GRAPH​, the calculator will draw the graphs.Now we are able to see that the graphs intersect at two points. They can be found by pressing 2ND​ and then CALC​.After selecting the intersect option, we need to choose left and right boundaries for one of the points. Finally, the calculator asks for a guess where the intersection point might be. After that, it will calculate the exact point for us. We will have to do this twice, once for each point.The solutions to the given system of equation are about (-1.88,2.35) and about (2.48,-4.64).