Exercise 25 Page 132

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+4x-5=7 ⇔ 2x^2+4x=12 Now let's divide each side by 2 so the coefficient of x^2 will be 1.

2x^2+4x=12

DivEqn

.LHS /2.=.RHS /2.

2x^2+4x/2=12/2

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Simplify left-hand side

WriteSumFrac

Write as a sum of fractions

2x^2/2+4x/2=12/2

MovePartNumRight

a* b/c=a/c* b

2/2x^2+4/2x=12/2

CalcQuot

Calculate quotient

x^2+2x=6

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

( b/2 )^2

Substitute

b= 2

( 2/2 )^2

CalcQuot

Calculate quotient

1

Next, we will add ( b2 )^2=1 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

x^2+2x=6

AddEqn

LHS+1=RHS+1

x^2+2x+ 1=6+ 1

FacPosPerfectSquare

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

(x+1)^2=6+1

AddTerms

Add terms

(x+1)^2=7

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x+1)^2)=sqrt(7)

CalcRoot

Calculate root

x+1=sqrt(7)

SubEqn

LHS-1=RHS-1

x=- 1± sqrt(7)

Both x=- 1+ sqrt(7) and x=-1-sqrt(7) are solutions of the equation. Now let's use a calculator to approximate each value of x. We see that rounded solutions are x=-3.7 and 1.7.