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###### Exercises

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Exercises 31 We will solve the equation and then compare our solution with the given work. To solve the equation using square roots, we have to start by isolating x on the left-hand side of the equation. 2x2−33=39LHS+33=RHS+332x2=72LHS/2=RHS/2x2=36 So far our solution is identical with the given work. Our next step will be taking the square root of each side. Don't forget that when solving an equation this way, we have to consider both the positive and the negative square roots. x2=36LHS=RHSx=±36Calculate rootx=±6 Therefore, the solutions to the equation are x=6 and x=-6. The given work is missing the negative square root of 36. | |

Exercises 32 Let's recall the formula for the volume of a rectangular prism. V=ℓwh Here ℓ is the length of the prism, w is the width of the prism, and h is its height, or in our case — its depth. We know that V=72000 and h=24. We also know that the length ℓ is two times the width w, so ℓ=2w. We will substitute these values into the formula. By doing so, we will obtain an equation that can be solved for w. V=ℓwhSubstitute V=72000,h=24,ℓ=2w72000=2w(w)(24)Multiply72000=48w2LHS/48=RHS/481500=w2LHS=RHS±1500=wRearrange equationw=±1500 The solutions to the equation are 1500 and -1500. Since w represents the width of the pond, it cannot be negative. Therefore, w=1500 and ℓ=21500. We can use a calculator to approximate our results. wℓ=1500≈38.7=21500≈77.5 The width is 1500≈38.7 inches and the length is 21500≈77.5 inches. | |

Exercises 33 We are given a function that represents the height h of the dropped sunglasses after x seconds. h=-16x2+24 The sunglasses hit the ground when h=0. Therefore, to determine how long it takes for the sunglasses to hit the ground, we have to determine the value of x when h=0. Let's substitute h=0 into the equation and simplify! h=-16x2+24h=00=-16x2+24LHS+16x2=RHS+16x216x2=24 LHS/16=RHS/16 LHS/16=RHS/16x2=1624ba=b/8a/8x2=23Calculate quotient x2=1.5LHS=RHSx=±1.5 The solutions to the equation are x=1.5 and x=-1.5. Since x represents time, it cannot be negative. Therefore, x=1.5≈1.2. It takes about 1.2 seconds for the sunglasses to hit the ground. | |

Exercises 34 Let's try to solve the given equation using square roots. To do so, we have to isolate x2 on one side of the equation. x2+4=0⇔x2=-4 Recall that a square of a real number cannot be negative. Therefore, the equation has no real solutions. Your cousin is correct. | |

Exercises 35 We will start by determining what must the area of the inner square be. Then we will use this information to find the side length x of the inner square.Area of the Inner Square We want the area of the inner square to be 25% of the total area of the rug. Let's determine the total area of the rug. We will use the formula for the area of a square. A=s2 Here s is the side length of the square. In our case s=6. Let's calculate the total area of the rug Ar. Ar=s2s=6Ar=62Calculate powerAr=36 Let Ai represent the area if the inner square. The area Ai must be equal to 25% of the area Ar. Ai=25%⋅Ar=25%⋅36 Let's simplify the obtained expression. Ai=25%⋅36a%=100aAi=10025⋅36ca⋅b=ca⋅bAi=10025⋅36ba=b/25a/25Ai=436Calculate quotientAi=9 The area of the inner square must be 9 square feet.Side Length of the Inner Square We have determined that the area of the inner square must be 9 square feet. Since x represents the side length of the inner square, we have the following equation. x2=9 We will solve the equation using square roots. However, we will only consider the positive square root because a negative solution does not make sense in this context. x=9=3 The side length of the inner square is 3 feet. | |

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Exercises 43 We want to find the x-coordinates of the points on the parabola which have y-coordinates of 9. Since these points are on the parabola, they must satisfy the equation of the parabola. Therefore, to find the x-coordinates, we will substitute y=9 into the given equation and solve it for x. y=21(x−2)2+1y=99=21(x−2)2+1 To solve this equation, we will rearrange it and isolate (x−2)2 on the left-hand side. 9=21(x−2)2+1Rearrange equation21(x−2)2+1=9LHS−1=RHS−121(x−2)2=8LHS⋅2=RHS⋅2(x−2)2=16 Next we will take a square root of each side of the equation. Remember about the positive and the negative square roots. (x−2)2=16LHS=RHSx−2=±16Calculate rootx−2=±4LHS+2=RHS+2x=2±4 The solutions to the equation are x=2+4=6 and x=2−4=-2. Therefore, the points on the parabola which have y-coordinates of 9 are (6,9) and (-2,9). | |

Exercises 44 aThe given equation cannot be simplified to the form x2=d. However, the left-hand side of the equation is a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored as shown below. a2+2ab+b2a2−2ab+b2=(a+b)2=(a−b)2 Let's identify the values of a and b for our trinomial and write it in the factored form. x2−12x+36=64⇕x2−2(x)(6)+62=64⇕(x−6)2=64 Next we will take the square root of each side! (x−6)2=64LHS=RHSx−6=±64Calculate rootx−6=±8LHS+6=RHS+6x=6±8 The solutions to the equation are x=6+8=14 and x=6−8=-2.bThe given equation cannot be simplified to the form x2=d. However, the left-hand side of the equation is a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored as shown below. a2+2ab+b2a2−2ab+b2=(a+b)2=(a−b)2 Let's identify the values of a and b for our trinomial and write it in the factored form. x2+14x+49=16⇕x2+2(x)(7)+72=16⇕(x+7)2=16 Next we will take the square root of each side! (x+7)2=16LHS=RHSx+7=±16Calculate rootx+7=±4LHS−7=RHS−7x=±4−7 The solutions to the equation are x=4−7=-3 and x=-4−7=-11. | |

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##### Other subchapters in Solving Quadratic Equations

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