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McGraw Hill Integrated II, 2012 View details

Study Guide and Review

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Exercise 83 Page 84

To solve this equation take the square root of each side.

-2 and - 12

Practice makes perfect

Notice that on the left hand side of the given equation we have a perfect square trinomial. To solve a quadratic equation in the form x^2=n, we will take the square root of each side. For any number n≥ 0, if x^2=n then x=±sqrt(n). Keeping this in mind, let's consider the given equation.

16d^2+40d+25=9

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(4d+5)^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

4d+5=±sqrt(9)

SubEqn

LHS-5=RHS-5

4d=-5±sqrt(9)

CalcRoot

Calculate root

4d=-5±3

DivEqn

.LHS /4.=.RHS /4.

d=-5±3/4

We can simplify this result into two separate roots.

d=- 5 ±3/4
d_1=- 5-3/4	d_2=- 5+3/4
d_1=- 8/4	d_2=- 2/4
d_1=- 2	d_2=- 1/2

We found that the solutions to the given equation are - 2 and - 12. To check our answer, we will graph the related function y=16d^2+40d+16 using a calculator. Note that in the calculator we will use the variable x instead of d.