Exercise 70 Page 672

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^2-6x+11=0 ⇕ x^2-6x=- 11In 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

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Simplify

MoveNegNumToFrac

Put minus sign in front of fraction

( - 6/2 )^2

CalcQuot

Calculate quotient

( - 3 )^2

NegBaseToPosPow

(- a)^2 = a^2

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=- 11

AddEqn

LHS+9=RHS+9

x^2-6x+ 9=- 11+ 9

FacNegPerfectSquare

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

(x-3)^2=- 11+9

AddTerms

Add terms

(x-3)^2=2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-3)^2)=sqrt(2)

CalcRoot

Calculate root

x-3=± sqrt(2)

AddEqn

LHS+3=RHS+3

x=3 ± sqrt(2)

Both x=3 + sqrt(2) and x=3 - sqrt(2) are solutions of the equation.