Exercise 8 Page 527

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

r ≈ - 9.24; r ≈ - 0.76

Practice makes perfect

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

r^2 + 10r = - 7
In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=10. Let's now calculate ( b2 )^2.

( b/2 )^2

b= 10

( 10/2 )^2

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Simplify

Calculate quotient

5^2

Calculate power

Next, we will add ( b2 )^2= 25 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

r^2 + 10r = - 7

LHS+25=RHS+25

r^2 + 10r + 25 = - 7 + 25

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

(r+5)^2 = - 7 + 25

Subtract term

(r+5)^2 = 18

sqrt(LHS)=sqrt(RHS)

sqrt((r+5)^2)=sqrt(18)

Calculate root

r+5 = ± 4.242640 ⋯

Round to 2 decimal place(s)

r+5 ≈ ± 4.24

LHS-5=RHS-5

r ≈ - 5 ± 4.24

Rounded to the nearest hundredth, both r ≈ - 9.24 and r ≈ - 0.76 are solutions of the equation.

Exercises p.527

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