Exercise 31 Page 159

We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so that all terms with x are on one side of the equation and the constant term is on the other. x^2+6x+9=16 ⇓ x^2+6x=7 In a quadratic expression, b is the linear coefficient. From the equation above, we know 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

▼

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

SplitIntoFactors

Split into factors

x^2+2x(3)+9=7+9

WritePow

Write as a power

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

The solutions for this equation are x=-3±4. Let's separate them into positive and negative cases.

x=-3±4
x_1=-3 + 4	x_2=-3 - 4
x_1=1	x_2=-7

We found that the solutions of the given equation are x_1=1 and x_2=- 7.