Exercise 7 Page 527

To complete the square, make sure all the variable terms are on one side of the equation.

x ≈ - 1.45; x ≈ 3.45

Practice makes perfect

We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation.

x^2-2x=5
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

b= - 2

( - 2/2 )^2

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Simplify

Calculate quotient

( - 1 )^2

Calculate power

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=5

LHS+1=RHS+1

x^2 - 2x + 1 = 5 + 1

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

(x-1)^2 = 5 + 1

Add terms

(x-1)^2 = 6

sqrt(LHS)=sqrt(RHS)

sqrt((x-1)^2)=sqrt(6)

Calculate root

x-1= ± 2.449489 ⋯

Round to 2 decimal place(s)

x-1 ≈ ± 2.45

LHS+1=RHS+1

x ≈ 1 ± 2.45

Rounded to the nearest hundredth, both x ≈ - 1.45 and x ≈ 3.45 are solutions of the equation.

Exercises p.527

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