Exercise 24 Page 132

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

-2.3 and 4.3

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. x^2-2x-10=0 ⇔ x^2-2=10In 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

▼

Simplify

ReduceFrac

a/b=.a /2./.b /2.

( 1/1 )^2

DivByOne

a/1=a

1^2

CalcPow

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=10

AddEqn

LHS+1=RHS+1

x^2-2x+ 1=10+ 1

FacNegPerfectSquare

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

(x-1)^2=10+1

AddTerms

Add terms

(x-1)^2=11

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-1)^2)=sqrt(11)

CalcRoot

Calculate root

x-1=± sqrt(11)

AddEqn

LHS+1=RHS+1

x=1± sqrt(11)

Both x=1+sqrt(11) and x=1- sqrt(11) are solutions of the equation. Now let's use a calculator to approximate each value of x. We see that rounded solutions are x=-2.3 and x=4.3.

Hints

Check the answer