Exercise 132 Page 308

Practice makes perfect

a We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with x are on one side of the equation and all constants are on the other side.

x^2+18x+32=0 ⇕ x^2+18x=- 32In a quadratic expression b is the linear coefficient. For the equation above, we have that b=18. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= 18

( 18/2 )^2

▼

Simplify

CalcQuot

Calculate quotient

9^2

CalcPow

Calculate power

Next, we will add ( b2 )^2=81 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

x^2+18x=- 32

AddEqn

LHS+81=RHS+81

x^2+18x+ 81=- 32+ 81

FacPosPerfectSquare

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

(x+9)^2=- 32+81

AddTerms

Add terms

(x+9)^2=49

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x+9)^2)=sqrt(49)

CalcRoot

Calculate root

x+9=± 7

SubEqn

LHS-14=RHS-14

x=- 9 ± 7

The solutions for this equation are x=- 9 ± 7. Let's separate them into the positive and negative cases.

x=- 9 ± 7
x_1=- 9 + 7	x_2=- 9 - 7
x_1=- 2	x_2=- 16

We found that the solutions of the given equation are x_1=- 2 and x_2=- 16.

b We want to solve the quadratic equation by completing the square. Note that the equation is already written in a form that all terms with x are on one side of the equation and all constants on the other side.

-36=x^2+12x ⇕ x^2+12x=-36In a quadratic expression b is the linear coefficient. For the equation above, we have that b=12. Let's now calculate ( b2 )^2.