Math 80/80B
Unit 8.2
SUPPLEMENT
SOLVING QUADRATIC EQUATIONS
BY THE QUADRATIC FORMULA
Completing the square involves a lot of work and is, therefore, not used very often in solving quadratic
equations. It is, however, very useful in finding the turning point or vertex of a parabola when more
advanced methods are not available.
A more practical method for solving quadratic equations, when factoring or root extraction cannot be
used, is the Quadratic Equation.
Quadratic Equation
x
b  b2  4ac
2a
is the quadratic equation and must be memorized.
The letters a, b, c are the same letters in the equation: ax  bx  c
2
Using the quadratic equation, one can solve any second-degree equation.
Example 1
Solve: 4 x  5x  7
2
1. Write the equation in standard form.
4 x2  5x  7  0
2. Identify the coefficients.
a  4, b  5, c  7
3. Substitute these values into the formula taking special care to distinguish between the
negative sign of the formula and those of the number.
x
  5 
 5  4  4  7 
2  4
2
4. Simply the right side.
x
5  25   112 
8
x
5  25  112
8
x
5  137
8
Therefore, the solutions are: x 
5  137
8
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
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PAGE 2
Example 2:
1 2
1
m m  0
3
2
Solve:
1
3
Solution: We could use the quadratic formula with a  , b  1, c 
1
. Instead, let’s
2
find a simpler, equivalent, standard-form equation whose coefficients are not fractions.
1. First we multiply both sides of the equation by 6 to clear the fractions.
1
1
6  m2  m    6  0
2
3
2m 2  6m  3  0
Simplify.
2. Identify the coefficients
a  2, b  6, c  3
3. Substitute these values into the formula taking special care to distinguish between the
negative sign of the formula and those of the number.
m
  6  
 6   4  2  3
2  2
2
4. Simply the right side.
m
m
6  36   24 
4
6  12
4


1
2 3 3
62 3
3 3
m


4
42
2
Therefore, the solutions are: m 
3 3
2
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
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PAGE 3
Using the Discriminant
x
b  b2  4ac
b2 - 4ac is called the discriminant because
2a
, the radicand
In the quadratic formula
when we know its value, we can discriminate among the possible number and type of solutions of a
quadratic equation. Possible values of the discriminant and their meanings are summarized next.
Discriminant
The following table relates the discriminant b2 - 4ac of a quadratic equation of the form
ax2  bx  c  0 with the number and type of solutions of the equation.
b2 - 4ac
Number and Type of Solutions
Two real solutions
Positive
One real solution
Zero
Two imaginary, not real solutions
Negative
The discriminant helps us determine the number and type of solutions of a quadratic,
ax2  bx  c  0 . The discriminant of ax2  bx  c  0 also tells us the number of
x-intercepts for the graph of f  x   ax 2  bx  c , or, equivalently, y  ax 2  bx  c .
b2 - 4ac  0
b2 - 4ac  0
b2 - 4ac  0
f(x) has two x-intercept
f(x) has one x-intercept
f(x) has no x-intercepts
Example 3: Use the discriminant to determine the number and type of solutions of
x2  2 x  3.
2
Solution: In x  2 x  3  0, a  1, b  2, c  3.
2
2
Thus, b  4ac  2  4 1 3   8
2
Since b  4ac is negative, this quadratic equation has two imaginary, not real solutions.
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
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PAGE 4
Example 4:
A pen is thrown upward from the top of a bridge. The stone’s height
in feet, above the water t seconds after the stone is thrown is given by the
2
function h  16t  14t  240
a. Find the maximum height of the stone.
b. Find the time it takes the stone to hit the water. Round to the nearest hundredth of a
second.
240 ft
1. UNDERSTAND. Read and reread the problem.
2. TRANSLATE. Since we want to know the maximum height of the stone for part a,
we need to use the vertex formula t 
3.
b
.
2a
2
Using the quadratic function h  16t  32t  240 , first we need to identify a and
b. a  16, b  32
t
32
2  16 
t  1sec
2
4. Substitute the time or x-value into the quadratic function h  16t  32t  240 to
find the maximum height of the stone.
h  16 1  14 1  240
2
h  238 ft
5.
To find the time it takes the stone to hit the water, we set h  0 .
0  16t 2  32t  240 and solve for t.
0  16  t 2  2t  15 
0   t  3 t  5 
t  3, t  5
t  5sec
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
*Exclude the negative answer
because negative time does not exist.
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PAGE 5
Problems:
Use the quadratic formula to solve each equation.
1.
x2  4 x  8  0
2.
p 2  12 p  14  0
3.
3y  6 y2  4
4.
6 x2  4  15x
5.
x2  4 x  4  0
6.
y 2  10 y  24  0
7.
1 2
x  x2  0
2
8.
3 2 1
1
y  y 0
5
4
2
9.
1 2
1
y  y
2
3
Use the discriminant to determine the number and types of solutions of each equation.
18.
10.
x2  6  0
11.
3x 2  4  6 x
12.
9 x  3x 2  5  0
13.
7 x  5  9 x2
14.
9 x2  2  5x
15.
3  2 x  10 x2  0
16.
5 x 2  3x  2  0
17. 4 x2  12 x  6  0
A stone is thrown upward from the top of a bridge. The stone’s height s  t  in feet, above the
2
water t seconds after the stone is thrown is given by the function s  t   16t  32t  280.
a. Find the maximum height of the stone.
b. Find the time it takes the stone to hit the water.
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
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PAGE 6
Answers:
1.
2  2 3
2.
6  5 2
3.
3  57
6
4.
15  321
12
5.
2
6.
4,  6
7.
1 5
8.
5  445 i
24
9.
3  15
3
10.
2 real solutions
11.
13. 2 real solutions 14. 2 complex solutions
17. 2 real solutions 18.
a) 296 ft
2 real solutions
12.
2 real solutions
15. 2 complex solutions 16.
2 real solutions
b)  5sec
[MATH80B/80 SUPPLEMENT: UNIT 8.2]
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No part of this work may be reproduced without the prior written consent of the Cerritos College Math Learning Center.