Exercise 32 Page 159

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 a are on one side of the equation and the constant term is on the other. - a^2-10a+25=25 ⇓ - a^2-10a=0 Now, let's multiply each side by -1 so the coefficient of a^2 will be 1.

- a^2-10a=0

MultEqn

LHS * (- 1)=RHS* (- 1)

-1(- a^2-10a)=0

Distr

Distribute -1

a^2+10a=0

In a quadratic expression, b is the linear coefficient. From the equation above, we know that b=10. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= 10

( 10/2 )^2

CalcQuot

Calculate quotient

5^2

CalcPow

Calculate power

25

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.

a^2+10a=0

AddEqn

LHS+25=RHS+25

a^2+10a+ 25=0+ 25

▼

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

SplitIntoFactors

Split into factors

a^2+2a(5)+5(5)=0+25

WritePow

Write as a power

a^2+2a(5)+5^2=0+25

FacPosPerfectSquare

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

(a+5)^2=0+25

IdPropAdd

Identity Property of Addition

(a+5)^2=25

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((a+5)^2)=sqrt(25)

SqrtPowToNumber

sqrt(a^2)=a

a+5=sqrt(25)

CalcRoot

Calculate root

a+5=± 5

SubEqn

LHS-5=RHS-5

x=- 5± 5

The solutions for this equation are x=-5±5. Let's separate them into positive and negative cases.

x=-5±5
x_1=-5 + 5	x_2=-5 - 5
x_1=0	x_2=-10

We found that the solutions of the given equation are x_1=0 and x_2=- 10.