Exercise 132 Page 308

Practice makes perfect

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.

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.