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McGraw Hill Integrated II, 2012 View details

Study Guide and Review

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Exercise 81 Page 84

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

-2 and 1

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.

4c^2+4c+1=9

FacPosPerfectSquare

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

(2c+1)^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

2c+1=±sqrt(9)

SubEqn

LHS-1=RHS-1

2c=-1±sqrt(9)

CalcRoot

Calculate root

2c=-1±3

DivEqn

.LHS /2.=.RHS /2.

c=-1±3/2

We can simplify this result into two separate roots.

c=- 1 ±3/2
c_1=- 1-3/2	c_2=- 1+3/2
c_1=- 4/2	c_2=2/2
c_1=- 2	c_2=1

We found that the solutions to the given equation are - 2 and 1. To check our answer, we will graph the related function y=4c^2+4c-8 using a calculator. Note that in the calculator we will use the variable x instead of c.