Exercise 29 Page 574

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. 2x^2+12x+20=34 ⇔ 2x^2+12x=14 Now let's divide each side by 2 so the coefficient of x^2 will be 1. 2x^2+12x=14 ⇓ x^2+6x=7 In a quadratic expression b is the linear coefficient. For the equation above, we have that b=6. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= 6

( 6/2 )^2

CalcQuot

Calculate quotient

3^2

CalcPow

Calculate power

9

Next, we will add ( b2 )^2=9 to both sides of our equation. Then we will factor the trinomial on the left-hand side and solve the equation.

x^2+6x=7

AddEqn

LHS+9=RHS+9

x^2+6x+ 9=7+ 9

FacPosPerfectSquare

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

(x+3)^2=7+9

AddTerms

Add terms

(x+3)^2=16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x+3)^2)=sqrt(16)

CalcRoot

Calculate root

x+3=± 4

SubEqn

LHS-3=RHS-3

x=- 3± 4

We conclude that the equation has two solutions. x_1&=- 3+4=1 x_2&=- 3-4=- 7