Chapter 8
Quadratic Equations and
Functions
§ 8.1
The Square Root Property and Completing
the Square
The Square Root Property
The Square Root Property
If u is an algebraic expression and d is a nonzero real
number, then u 2  d has exactly two solutions:
If u 2  d , then u  d or u   d .
Equivalently,
If u 2  d , then u   d .
Blitzer, Intermediate Algebra, 4e – Slide #3
Using the Square Root Property
EXAMPLE
Solve: 4 x 2  49.
SOLUTION
To apply the square root property, we need a squared expression
by itself on one side of the equation. We can get x 2 by itself if
we divide both sides by 4.
4 x 2  49
This is the given equation.
x2 
49
4
Divide both sides by 4.
49
x
4
49
7
x
   3.5
2
4
Apply the square root property.
Simplify.
Blitzer, Intermediate Algebra, 4e – Slide #4
Using the Square Root Property
CONTINUED
Check 3.5:
Check -3.5:
4 x 2  49
4 x 2  49
43.5 ? 49
4 3.5 ? 49
4 12.25 ? 49
4 12.25 ? 49
2
49  49 true
2
49  49 true
The solutions are 3.5 and -3.5. The solution set is {3.5,-3.5}.
Blitzer, Intermediate Algebra, 4e – Slide #5
Using the Square Root Property
EXAMPLE
Solve: 4 x 2  49  0.
SOLUTION
To solve by the square root property , we isolate the squared
expression on one side of the equation.
4 x 2  49  0
This is the given equation.
Subtract 49 from both sides.
4 x 2  49
49
2
x 
4
x 
Divide both sides by 4.
49
4
Apply the square root property.
Blitzer, Intermediate Algebra, 4e – Slide #6
Using the Square Root Property
CONTINUED
49
1
4
49
7
x
i   i  3.5i
2
4
Check 3.5i:
4 x 2  49  0
x
1  i
Check -3.5i:
4 x 2  49  0
43.5i   49 ? 0
2
4 3.5i   49 ? 0
4 12.25i 2  49 ? 0
4 12.25i 2  49 ? 0
2
4 12.251  49 ? 0
4 12.251  49 ? 0
0  0 true
0  0 true
The solutions are 3.5i and -3.5i. The solution set is {3.5i,-3.5i}.
Blitzer, Intermediate Algebra, 4e – Slide #7
Using the Square Root Property
EXAMPLE
Solve by the square root property: 3x  22  36.
SOLUTION
3x  2  36
2
This is the given equation.
Divide both sides by 3.
x  22  12
x  2  12 or x  2   12
x  2  12 or x  2  12
x  2  4  3 or x  2  4  3
x  2  2 3 or x  2  2 3
Apply the square root property.
Subtract 2 from both sides of each
equation.
Rewrite radicands.
Simplify.
Blitzer, Intermediate Algebra, 4e – Slide #8
Using the Square Root Property
CONTINUED
Check  2  2 3:
Check  2  2 3:
3x  2  36
3x  2  36
2


32 3  ? 36
3  2   3  ? 36
2
3  2  2 3  2  36
?
2
2
2
3  4  3 ? 36
36  36 true
2


3 2 3  ? 36
3 2   3  ? 36
2
3  2  2 3  2 ? 36
2
2
2
3  4  3 ? 36
36  36 true


The solutions are  2  2 3. The solution set is  2  2 3 .
Blitzer, Intermediate Algebra, 4e – Slide #9
Completing the Square
Completing the Square
2
b
If x 2  bx is a binomial, then by adding  , which is the
2
square of half the coefficient of x, a perfect square
trinomial will result. That is,
2
2
b
b 
x  bx      x   .
2
 2 
2
Blitzer, Intermediate Algebra, 4e – Slide #10
Completing the Square
EXAMPLE
What term should be added to the binomial so that it becomes a
perfect square trinomial? Write and factor the trinomial.
x 2  10x
SOLUTION
x 2  10x
2
 10 
2
Add 
   5 .
 2 
x2 10x  25  x  5
2
Add 25 to complete
the square.
Blitzer, Intermediate Algebra, 4e – Slide #11
Completing the Square
EXAMPLE
Solve by completing the square: 2 x 2  5 x  3  0.
SOLUTION
2 x 2  5x  3  0
5
3
2
x  x 0
2
2
x2 
5
3
x
2
2
5
25 3 25
x  x
 
2
16 2 16
2
This is the given equation.
Divide both sides by 2.
Add 3/2 to both sides.
Complete the square: Half of 5/2 is
2
5
5
25
 
, and    .
4
 4  16
Blitzer, Intermediate Algebra, 4e – Slide #12
Completing the Square
CONTINUED
2
5
24 25 49



x  
4  16 16 25

5
49
x 
4
25
5
7
x 
4
5
7 5
7 5
x   or x   
5 4
5 4
x
28 25
28 25

or x   
20 20
20 20
Factor and simplify.
Apply the square root property.
49
49 7


25
25 5
Split into two equations and
subtract 5/4 from both sides of
both equations.
Get common denominators.
Blitzer, Intermediate Algebra, 4e – Slide #13
Completing the Square
CONTINUED
x
3
53
or x  
20
20
Simplify.
3
53
3 53
The solutions are
and  , and the solution set is 
, .

20
20
 20 20
Blitzer, Intermediate Algebra, 4e – Slide #14
Completing the Square in Application
A Formula for Compound Interest
Suppose that an amount of money, P, is invested at rate r,
compounded annually. In t years, the amount, A, or
balance, in the account is given by the formula
A  P1  r  .
t
Blitzer, Intermediate Algebra, 4e – Slide #15
Completing the Square in Application
EXAMPLE
Use the compound interest formula, A  P1  r t , to find the
annual interest rate r.
In 2 years, an investment of $80,000 grows to $101,250.
SOLUTION
We are given that
P (the amount invested) = $80,000
t (the time of the investment) = 2 years
A (the amount, or balance, in the account) = $101,250.
Blitzer, Intermediate Algebra, 4e – Slide #16
Completing the Square in Application
CONTINUED
We are asked to find the annual interest rate, r. We substitute
the three given values into the compound interest formula and
solve for r.
A  P1  r 
t
101,250  80,0001  r 
101,250
2
 1  r 
80,000
81
2
 1  r 
64
81
81
1 r 
or 1  r  
64
64
2
Use the compound interest formula.
Substitute the given values.
Divide both sides by 80,000.
Simplify the fraction.
Apply the square root property.
Blitzer, Intermediate Algebra, 4e – Slide #17
Completing the Square in Application
CONTINUED
9
9
or 1  r  
8
8
9
9
r   1 or r    1
8
8
1 r 
r
1
17
or r  
8
8
81
81 9


64
64 8
Subtract 1 from both sides.
Simplify.
Because the interest rate cannot be negative, we reject -17/8.
Thus, the annual interest rate is 1/8 = 0.125 = 12.5%.
We can check this answer using the formula A  P1  r . If
$80,000 is invested for 2 years at 12.5% interest, compounded
annually, the balance in the account is
t
Blitzer, Intermediate Algebra, 4e – Slide #18
Completing the Square in Application
CONTINUED
A  $80,0001  0.125  $80,0001.125  $101,250.
2
2
Because this is precisely the amount given by the problem’s
conditions, the annual interest rate is, indeed, 12.5%
compounded annually.
Blitzer, Intermediate Algebra, 4e – Slide #19
Isosceles Right Triangles
Lengths Within Isosceles Right Triangles
The length of the hypotenuse of an isosceles right triangle is
the length of a leg times 2.
a 2
a
a
Blitzer, Intermediate Algebra, 4e – Slide #20
Isosceles Right Triangles in Application
EXAMPLE
A supporting wire is to be attached to the top of a 70-foot
antenna. If the wire must be anchored 70 feet from the base of
the antenna, what length of wire is required?
SOLUTION
Since the supporting wire, the antenna, and a line on the
ground between the base of the antenna and the base of the
supporting wire form an isosceles right triangle as shown
below,
antenna
supporting wire
Blitzer, Intermediate Algebra, 4e – Slide #21
Isosceles Right Triangles in Application
CONTINUED
this implies that the diagram can be represented as follows,
using the “lengths within isosceles right triangles” principle.
antenna
70
ft.
supporting
2 ft. wire
70
70 ft.
Therefore the supporting wire must be 70 2 feet which equals
about 99 ft.
Blitzer, Intermediate Algebra, 4e – Slide #22