body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Mid-Chapter Quiz

Continue to next subchapter

Exercise 23 Page 132

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

-0.6 and 3.4

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+4x+2=0 ⇔ x^2+4x=-2In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=4. Let's now calculate ( b2 )^2.

( b/2 )^2

b= 4

( 4/2 )^2

▼

Simplify

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

2^2

Calculate power

4

Next, we will add ( b2 )^2=4 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

x^2+4x=-2

LHS+4=RHS+4

x^2+4x+4=2

FacPosPerfectSquare

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

(x+2)^2=2

sqrt(LHS)=sqrt(RHS)

sqrt((x+2)^2)=sqrt(2)

Calculate root

x+2=± sqrt(2)

LHS-2=RHS-2

x=- 2± sqrt(2)

Both x=- 2+ sqrt(2) and x=- 2- sqrt(2) are solutions of the equation. Now let's use a calculator to approximate each value of x. We see that rounded solutions are x=-0.6 and x=3.4.