Beginning Algebra, 11th Edition

NOTE The solutions given in Example 7are exact. In applications, dec-

imal solutions are more appropriate. Using the square root key of a calculator yields

Approximating the two solutions gives

xL1.449 and xL-3.449.

26 L2.449.

- 1  26

SECTION 9.2 Solving Quadratic Equations by Completing the Square 565

Complete solution available
on the Video Resources on DVD

9.2 EXERCISES

Complete each trinomial so that it is a perfect square. Then factor the trinomial. See

Example 1.

1. 
   
   
   
   2. 
      
      
      
      3. 
         
      
      
      
      
   
   
   
   


2. 
   
   
   
   5. 
      
      
      
      6. 
         
      
      
      
      
   
   
   
   


3. 
   
   
   
   8. 9.Concept Check Which step is an appropriate way to begin solving the quadratic equa-
      tion by completing the square?
      A.Add 4 to each side of the equation. B.Factor the left side as.
      C.Factor the left side as. D.Divide each side by 2.
      10.Concept Check In Example 3 of Section 6.5, we solved the quadratic equation
      by factoring. If we were to solve by completing the square, would we
      get the same solution set, E^52 , 4F?
   
   
   
   

4 p^2 + 40 = 26 p

x 12 x- 42

2 x 1 x- 22

2 x^2 - 4 x= 9

p^2 - 5 p+ x^2 + 3 x+

a^2 - 32 a+ x^2 + 2 x+ m^2 - 2 m+

x^2 + 10 x+ x^2 + 16 x+ z^2 - 20 z+

NOW TRY
EXERCISE 8
At what times will the ball in
Example 8be 28 ft above the
ground?

NOW TRY ANSWER
8.0.5 sec and 3.5 sec

OBJECTIVE 4 Solve applied problems that require quadratic equations.

Solving a Velocity Problem

If a ball is projected into the air from ground level with an initial velocity of 64 ft per

sec, its altitude (height) sin feet in tseconds is given by the formula

At what times will the ball be 48 ft above the ground?

Since srepresents the height, we let in the formula and solve this equa-

tion for the time tby completing the square.

Let

Divide by

Interchange the sides.

Add

Factor. Add.

or Square root property

or Add 2.

The ball reaches a height of 48 ft twice, once on the way up and again on the way

down. It takes 1 sec to reach 48 ft on the way up, and then after 3 sec, the ball reaches

48 ft again on the way down. NOW TRY

t= 3 t= 1

t- 2 = 1 t - 2 =- 1

1 t- 222 = 1

t^2 - 4 t+ 4 =- 3 + 4 C 21 1 - 42 D^2 = 1 - 222 =4.

t^2 - 4 t=- 3

• 3 =t^2 - 4 t -16.

48 =- 16 t^2 + 64 t s=48.

s= 48

s=- 16 t^2 + 64 t.

EXAMPLE 8