Algebra
Module A62
Quadratic
Equations - 2
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED November, 2008
Quadratic Equations - 2
Statement of Prerequisite Skills
Complete all previous TLM modules before completing this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
The quadratic equation is one of the most common equations encountered in the
technologies. Its applications include projectile motion, architectural problems, electrical
circuits, problems in mechanics, and design. This unit will introduce the student to many
different strategies for manipulating and solving quadratic equations.
Learning Outcome
When you complete this module you will be able to…
Apply additional theory on quadratic equations.
Learning Objectives
1. Determine the nature of the roots of a quadratic equation by examining the value
of the discriminant.
2. Determine the quadratic equation given the roots of the equation.
3. Determine the quadratic equation given the sum and product of the roots.
Connection Activity
Recall the quadratic equation:
−b ± b 2 − 4ac
2a
We used this formula to solve for the roots of a quadratic equation in previous modules.
A close examination of the discriminant ( b 2 − 4ac ) will yield some useful information
about the quadratic in question. This module will guide you through using parts of a
quadratic such as the discriminant and the roots to provide information about quadratics.
x=
1
Module A62 − More on Quadratics - 2
OBJECTIVE ONE
When you complete this objective you will be able to…
Determine the nature of the roots of a quadratic equation by examining the value of the
discriminant.
Exploration Activity
Discriminant
In the quadratic formula x =
−b ± b 2 − 4ac
, the quantity b 2 − 4ac is called the
2a
discriminant.
b 2 − 4ac helps us determine the type of roots we will get when we solve a quadratic
equation. The discriminant does not include the radical sign.
We shall shorten the word discriminant to the word disc for convenience. There are three
types of discriminant that we are going to study.
TYPE I:
If b 2 − 4ac > 0, the roots are real and unequal.
Example 1
Determine the discriminant for the following quadratic equation. Then use this to solve
the equation.
x2 − 2 x − 3 = 0
SOLUTION:
In this equation a = 1, b = −2, c = −3.
Substituting into b 2 − 4ac we get:
b 2 − 4 ac
= (−2)2 − 4(1)(−3)
= 16
Since 16 > 0, i.e, the discriminant > 0, the roots are real and unequal. Knowing
this, solve the quadratic:
x2 − 2 x − 3 = 0
(x − 3)(x + 1) = 0
x = 3, and x = −1
2
Module A62 − More on Quadratics - 2
Check in x 2 − 2 x − 3 = 0 by substituting.
For x = 3:
32 − 2(3) − 3 = 0
0 = 0 so 3 works.
For x = −1:
(−1)2 − 2(−1) −3 = 0
0 = 0 so −1 works.
TYPE II
If b 2 − 4ac = 0 , the roots are real and equal.
Example 2
Determine the discriminant for the following quadratic equation. Then use this to solve
the equation.
4x2 + 28x + 49 = 0
SOLUTION:
Here a = 4, b = 28, c = 49
b 2 − 4ac = 282 − 4(4)(49) = 0
Since the discriminant = 0, the roots are real and equal.
Knowing that the roots are real and equal makes the equation easier to factor:
4x2 + 28x + 49 = 0
(2x + 7)(2x + 7) = 0
7 7
x = − ,−
2 2
You should check these answers by substitution.
TYPE III
If b 2 − 4ac < 0 , the roots are imaginary because we cannot take the square root of a
−b ± negative number
negative number. Looking at the quadratic formula: x =
2a
3
Module A62 − More on Quadratics - 2
Example 3
Determine the discriminant for the following quadratic equation. Use this to solve the
equation.
3x2 − 2x + 5 = 0
SOLUTION:
Here a = 3, b = −2, c = 5
b 2 − 4ac = (−2)2 − 4(3)(5)
= −56
Since the discriminant is less than zero, the roots are imaginary. Don’t try to
solve by factoring. Instead go directly to the quadratic formula.
−b ± b 2 − 4ac
2a
2 ± −56
=
6
2 + 7.48i
2 − 7.48i
and
=
6
6
= 0.33 + 1.25i and 0.33 − 1.25i
x=
Check your answer.
SUMMARY
The discriminant is a number that will assist you in solving a quadratic equation:
If the discriminant is 0, then the roots are real and equal.
If the discriminant is > 0, then the roots are real and unequal.
If the discriminant is < 0, then the roots are imaginary.
4
Module A62 − More on Quadratics - 2
Experiential Activity One
Determine the discriminant for each of the following quadratic equations. Use this to
help you solve the given equations.
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
x2 + x − 3 = 0
x2 – 3x + 1 = 0
2t2 + 3t + 1 = 0
x2 – 2x = 1
3t2 + 6t = −2
x2 x
− =1
6 3
x2 + 25 = 12x
y2 + 2y + 2 = 0
2x2 – 7x + 4 = 0
2t2 + 10t = −15
4x2 – 9 = 0
15 + 4x – 32x2 = 0
1 2 1
1
x + x− =0
4
3
12
3 2
1
x −x=
2
3
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
24.
26.
28.
x2 + x − 4 = 0
x2 – 5x + 5 = 0
2z2 + z – 2 = 0
x2 − 2x = 2
3x2 + 6x + 1 = 0
x2 x 1
− + =0
12 3 12
y2 + 23 = 10y
x2 – 2x + 2 = 0
3x2 – 5x = 4
2t2 + 7 = 4t
x2 – 6x = 0
4x2 – 12x = 7 Show Me.
1 2 1
1
x − x− =0
4
3
12
2 2 4
1
x − x=
3
9
3
Experiential Activity One Answers
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
13
5
1
8
12
0.7778
44
−4
17
−20
144
1936
0.1944
3
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
24.
26.
28.
17
5
17
12
24
0.0833
8
−4
73
−40
36
256
0.1944
1.0864
5
Module A62 − More on Quadratics - 2
OBJECTIVE TWO
When you complete this objective you will be able to…
Determine the quadratic equation given the roots of the equation.
Exploration Activity
Example 1
Determine the quadratic equation given the roots as:
x = −6, 1
then
x = −6, and x = 1
SOLUTION:
a) Determine factors
- move everything to the Left Hand Side (LHS)
x + 6 = 0 and x – 1 = 0
b) Multiply the factors.
( x + 6 )( x − 1) = 0
x2 − x + 6 x − 6 = 0
c) Simplify
x2 + 5x − 6 = 0
d) Check by substitutions
substitute the roots from above in and solve.
for x = –6
( −6 )
2
+ 5 ( −6 ) − 6 = 0
36 − 30 − 6 = 0
0=0
Both Checks Work !!
for x = 1
(1)
2
+ 5 (1) − 6 = 0
1+ 5 − 6 = 0
0=0
6
Module A62 − More on Quadratics - 2
Example 2
Determine the quadratic equation given the roots are:
x=
5
1
,−
3
2
Find the quadratic for
x=
5
1
and x = −
3
2
SOLUTION:
a) Determine factors
- clear denominators first.
3x = 5 and 2x = –1
- move everything to the LHS
3x – 5 = 0 and 2x +1 = 0
b) Multiply the factors.
( 3x − 5)( 2 x + 1) = 0
6 x 2 + 3 x − 10 x − 5 = 0
c) Simplify
6x2 − 7 x − 5 = 0
d) Check by substitutions
Complete this on your own.
7
Module A62 − More on Quadratics - 2
Example 3
Determine the quadratic equation given the roots are:
x = 3± 5
Find the quadratic for
x = 3 + 5 and x = 3 − 5
SOLUTION:
a) Determine factors
x = 3+ 5
x = 3− 5
x−3= 5
x−3= − 5
x −3− 5 = 0
x −3+ 5 = 0
b) Multiply the factors.
( x − 3 − 5 )( x − 3 + 5 ) = 0
x 2 − 3x + x 5 − 3x + 9 − 3 5 − 5 = 0
c) Simplify
x2 − 6 x + 4 = 0
d) Check by substitutions
Complete this on your own.
8
Module A62 − More on Quadratics - 2
Experiential Activity Two
Determine the quadratic equation for each of the following pairs of given roots.
1. x = 1, 3
3. x = 2, 4
5. x = 4, −4
5 5
7. x = , −
2 2
9. x = −4, −6
2. x = −2, −3
4. x = −6, 2
6. x = 0, 5
11. x = 4 ± 3
12. x = 1 ± 2
8. x = 1, 2
10. x = 1 ± 7 Show Me.
Experiential Activity Two Answers
1.
3.
5.
7.
9.
11.
x2 – 4x + 3 = 0
x2 – 6x + 8 = 0
x2 – 16 = 0
x2 – 6.250 = 0
x2 + 10x + 24 = 0
x2 – 8x + 13 = 0
2.
4.
6.
8.
10.
12.
x2 + 5x + 6 = 0
x2 + 4x – 12 = 0
x2 – 5x = 0
x2 – 3x + 2 = 0
x2 – 2x – 6 = 0
x2 – 2x – 1 = 0
9
Module A62 − More on Quadratics - 2
OBJECTIVE THREE
When you complete this objective you will be able to…
Determine the quadratic equation given the sum and product of the roots.
Exploration Activity
A quick way to determine an equation from the roots is to use the sum and product
of the roots.
Determining an Equation from the sum and product of the roots
Given the general quadratic equation: ax2 + bx + c = 0.
If we know the roots of the equation, coefficient b is the negative of the sum of
the roots and coefficient c is the product of the roots.
if the roots are r1 and r2 then
coefficient b is: – (r1 + r2) Å the neg. of the sum of the roots.
coefficient c is: r1 ⋅ r2 Å the product of the roots.
The equation can therefore be determined from the roots using the following:
x 2 − ( r1 + r2 ) x + r1r2 = 0
or
x 2 + ( negative of sum of roots ) x + product of roots = 0
Example 1
Construct the quadratic equation which has roots of 2 and 3.
SOLUTION:
Find the negative of the sum of the roots: – (r1 + r2)
= – (r1 + r2)
= – (2 + 3)
=–5
Therefore coefficient b is –5
Find the product of the roots: r1 ⋅ r2
= r1 ⋅ r2
=2⋅3
=6
Therefore coefficient c is 6
Substitute into x 2 + bx + c = 0
x2 − 5x + 6 = 0
10
Module A62 − More on Quadratics - 2
Example 2
Construct the quadratic equation which has roots of –3 and 4.
SOLUTION:
Find the negative of the sum of the roots: – (r1 + r2)
= – (r1 + r2)
= – (–3 + 4)
= –1
Therefore coefficient b is –1
Find the product of the roots: r1 ⋅ r2
= r1 ⋅ r2
= –3 ⋅ 4
= –12
Therefore coefficient c is 12
Substitute into x 2 + bx + c = 0
x 2 − 1x − 12 = 0
Example 3
Construct the quadratic equation which has roots of
3
2
and –2.
SOLUTION:
Find the negative of the sum of the roots: – (r1 + r2)
= – (r1 + r2)
= − ( 32 + −2 )
= − ( 32 − 42 )
= − ( − 12 )
= 12
Therefore coefficient b is
1
2
Find the product of the roots: r1 ⋅ r2
= r1 ⋅ r2
= 32 ⋅ −2
= –3
Therefore coefficient c is –3
Substitute into x 2 + bx + c = 0
x 2 + 12 x − 3 = 0
Clear the fractions by multiplying by the Lowest Common Denominator (LCD)
2 x2 + x − 6 = 0
11
Module A62 − More on Quadratics - 2
Experiential Activity Three
Find the quadratic equation given the sum and product of the roots.
SUM
PRODUCT
1.
7
6
2.
−3
−1.5
3.
5
−3
5/3
0
4.
5.
0
1.5
⎛m n ⎞
−⎜ + ⎟
1
6.
⎝ n m⎠
Find the equation given the roots by first determining the sum and products of the roots.
7. x = 1, 4
8. x = –3, 7
2 1
9. x = , −
3 2
10. x = 2 + 3, 2 − 3
Experiential Activity Three Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
x2 – 7x + 6 = 0
2x2 + 6x – 3 = 0
x2 + 3x + 5 = 0
3x2 – 5x = 0
2x2 + 3 = 0
mnx2 + (m2 + n2)x + mn = 0
x2 − 5x + 4 = 0
x 2 − 4 x − 21 = 0
6x2 − x − 2 = 0
x2 − 4x + 1 = 0
Show Me.
Practical Application Activity
Complete the Quadratics Equations - 2 assignment in TLM.
Summary
This module presents the student with additional theory on quadratic equations.
12
Module A62 − More on Quadratics - 2