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Unit 10: Quadratic Equations
Lesson 8: Using the Discriminant
The discriminant is a very useful tool when working with quadratic equations. The discriminant tells
us what kinds of solutions to expect when solving quadratic equations. Let’s take a look!
The Quadratic Formula
=
− ± √ − 4
2
The expression under the radical in the quadratic formula is called the discriminant.
The Discriminant
________________________
The discriminant indicates the number and type if solutions to a quadratic
equation if the original equation has integer coefficients.
The following table shows the relationship between the discriminant and the
type of solutions for the equation.
If the discriminant (b2 – 4ac) is:
Then the equation will have:
Negative
No real solutions
(There are no x-intercepts. The
graph does not cross the x-axis)
One rational solution
(This equation can be factored)
Zero
A positive number that is a perfect
square
Two rational solutions
(This equation can be factored)
A positive number that is not a perfect
square
Two irrational solutions
(You must use the quadratic
formula)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
Example 1
Use the discriminant to determine the number and type of solutions. If possible, solve the equation.
6x2 – 25x + 24 = 0
Step 1: The discriminant:
This equation has _______ solutions. They are __________________ so I will
_______________________________________________________________.
Step 2: Solve
Example 2
Use the discriminant to determine the number and type of solutions. If possible, solve the equation.
6x2 + x = 3
Step 1: The discriminant:
This equation has _______ solutions. They are __________________ so I will
_______________________________________________________________.
Step 2: Solve
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
Example 3
s Use the discriminant to determine the number and type of solutions. If possible, solve the equation.
2x2 = 2x - 5
Step 1: The discriminant:
This equation has _______ solutions. They are __________________ so I will
_______________________________________________________________.
Step 2: Solve
Example 4
Use the discriminant to determine the number and type of solutions. If possible, solve the equation.
4x2 – 16x + 16 = 0
Step 1: The discriminant:
This equation has _______ solutions. They are __________________ so I will
_______________________________________________________________.
Step 2: Solve
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
Lesson 8: Using the Discriminant
Directions: For each equation below, use the discriminant to determine the best method for solving
the equation. Then give all possible solutions.
1.
12x2 – 2x - 2 = 0
2.
y2 – 3y = -8
3.
3x2 = 4x + 5
4.
4r2 + 8r+ 4 = 0
5.
36x2 – 24x = -4
6.
5w2 + 3w – 6 = 0
7.
4p2 – 2p – 6 = 0
8.
9x2 = -6x – 1
9.
2a2 + 6a = 11
10.
2x2 – 2x + 10 = 0
Directions: For each equation below, use the discriminant to determine
the best method for solving the equation. Then give all possible
solutions.
1.
4x2 – 32x + 64 = 0
2.
3x2 – x = -5
(3 points each)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
Lesson 8: Using the Discriminant – Answer Key
Directions: For each equation below, use the discriminant to determine the best method for solving
the equation. Then give all possible solutions.
1.
12x2 – 2x - 2 = 0
This equation is already set equal to 0, so the values are as follows:
A = 12
B = -2
C =-2
The discriminant is: b2 – 4ac
(-2)2 – 4(12)(-2) = 100 The discriminant is a perfect square, therefore, the best
method is factoring. I will first factor out a 2.
2(6x2 – x – 1) = 0
Now factor inside of the parenthesis:
(3x +1)(2x-1) = 0
2.
3x+1 =0
3x+1-1= 0-1
3x/3 = -1/3
x= -1/3
2x-1 = 0
2x-1+1 = 0+1
2x/2 = ½
x=½
These are the x-intercepts
y2 – 3y = -8
This equation must be set equal to 0 so we must add 8 to both sides.
y2 – 3y +8 = -8+8
y2 – 3y +8 = 0. Now the values are:
A=1
B = -3
C =8
The discriminant is: b2 – 4ac
(-3)2 – 4(1)(8) = -23 The discriminant is negative, therefore, the solutions are not
real numbers. Therefore, we will stop here. This parabola when graphed will not cross the x-axis.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
3x2 = 4x + 5
3.
First we must set the equation equal to 0 by subtracting 3x2 from both sides.
3x2 – 3x2 = -3x2+4x+5
0 = -3x2 +4x+5
A = -3
or
- 3x2 +4x+5 = 0
B=4
C =-5
The discriminant is: b2 – 4ac
(4)2 – 4(-3)(5) = 76 The discriminant is positive, but not a perfect square.
Therefore, it’s not factorable and we must use the quadratic formula to solve.
=
−4 ± 4 − 4−35
2−3
=
−4 ± √76
−6
x = -.79 and
4.
x = 2.12
These are the x-intercepts
4r2 + 8r+ 4 = 0
This is set equal to 0, so our values are as follows:
A=4
B=8
C =4
The discriminant is: b2 – 4ac
(8)2 – 4(4)(4) = 0 The discriminant is 0, therefore, there is only one real solution.
This equation should be factorable and it will follow one of the special rules for binomials.
First factor out the GCF, which is 4.
4(r2 +2r +1) = 0
Now factor inside of the parenthesis.
4(r+1)(r+1) = 0
or 4(r+1)2 = 0
r+1 = 0
r+1 – 1 = 0-1
r = -1
(Both factors would be r = -1; therefore, there is only one real solution. This parabola has a vertex on the
x-axis and that’s why there’s only one x-intercept.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
5. 36x2 – 24x = -4
First we must set this equation equal to 0 by adding 4 to both sides.
36x2 – 24x +4 = -4 +4
36x2 – 24x +4 = 0
A = 36
B = -24
C =4
The discriminant is: b2 – 4ac
(-24)2 – 4(36)(4) = 0 The discriminant is 0, therefore, there is only one real
solution. This equation should be factorable and it will follow one of the special rules for
binomials.
First factor out the GCF, which is 4.
4(9r2 -6r +1) = 0
(Notice how the “A” term and “c” term are both perfect squares?)
Now factor inside of the parenthesis.
4(3r+1)(3r+1)= 0
or 4(3r+1)2 = 0
3r+1 = 0
3r+1 – 1 = 0-1
3r = -1
3r/3 = -1/3
r = -1/3
(Both factors would be r = -1/3; therefore, there is only one real solution. This parabola has a vertex on
the x-axis and that’s why there’s only one x-intercept.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
6. 5w2 + 3w – 6 = 0
This4.equation is already set equal to 0. Therefore,
A=5
B=3
C = -6
The discriminant is: b2 – 4ac
(3)2 – 4(5)(-6) = 129 The discriminant is positive, but not a perfect square.
Therefore, it’s not factorable and we must use the quadratic formula to solve.
=
−3 ± 3 − 45−6
25
=
−3 ± √129
10
x = .84 and
x = -1.44
These are the x-intercepts
7. 4p2 – 2p – 6 = 0
This equation is already set equal to 0, so the values are as follows:
A=4
B = -2
C = -6
The discriminant is: b2 – 4ac
(-2)2 – 4(4)(-6) = 100 The discriminant is a perfect square, therefore, the best
method is factoring. I will first factor out a 2.
2(2p2 – p – 3) = 0
Now factor inside of the parenthesis:
(2p -3)(p+1) = 0
2p-3 = 0
2p-3+3= 0+3
2p=3
2p/2 = 3/2
p = 3/2
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
p+1 = 0
p+1-1=0-1
p= -1
p = -1
These are the x-intercepts
Unit 10: Quadratic Equations
8. 9x2 = -6x – 1
First we must set this equation equal to 0 by subtracting 9x2 to both sides.
9x2 – 9x2 = -9x2 – 6x – 1
0 = -9x2 – 6x – 1
A = -9
or
-9x2 – 6x – 1 = 0
B = -6
C =-1
The discriminant is: b2 – 4ac
(-6)2 – 4(-9)(-1) = 0 The discriminant is 0, therefore, there is only one real
solution. This equation should be factorable and it will follow one of the special rules for
binomials.
Let’s multiply by -1 to make the lead coefficient positive.
-1(-9x2 – 6x – 1) = 0(-1)
9x2 + 6x +1 = 0
(Notice how the “A” term and “c” term are both perfect squares?)
(3x +1)(3x+1) = 0
or (3x+1)2 = 0
3x+1 = 0
3x+1 – 1 = 0-1
3x = -1
3x/3 = -1/3
x = -1/3
(Both factors would be x = -1/3; therefore, there is only one real solution. This parabola has a vertex on
the x-axis and that’s why there’s only one x-intercept.
9. 2a2 + 6a = 11
This equation must be set equal to 0 by subtracting 11 from both sides.
2a2 +6a – 11 = 11-11
or
A=2
C = -11
B=6
2a2 +6a – 11 = 0
The discriminant is: b2 – 4ac
(6)2 – 4(2)(-11) = 124 The discriminant is positive, but not a perfect square.
Therefore, it’s not factorable and we must use the quadratic formula to solve.
=
−6 ± 6 − 42−11
22
=
−6 ± √124
4
x = 1.28 and
x = -4.28
These are the x-intercepts
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
10. 2x2 – 2x + 10 = 0
This equation is set equal to 0, therefore,
A=2
B = -2
C =10
The discriminant is: b2 – 4ac
(-2)2 – 4(2)(10) = -76 The discriminant is negative, therefore, the solutions are not
real numbers. Therefore, we will stop here. This parabola when graphed will not cross the x-axis.
Directions: For each equation below, use the discriminant to determine
the best method for solving the equation. Then give all possible
solutions.
(3 points each)
1. 4x2 – 32x + 64 = 0
This equation is set equal to 0; therefore,
A=4
B = -32
C =64
The discriminant is: b2 – 4ac
(-32)2 – 4(4)(64) = 0 The discriminant is 0, therefore, there is only one real
solution. This equation should be factorable and it will follow one of the special rules for
binomials.
4x2 -32x +64 = 0
(Notice how the “A” term and “c” term are both perfect squares?)
(2x – 8)(2x-8) = 0
or (2x-8)2 = 0
Set one of the factors equal to 0.
2x-8 = 0
2x-8+8 = 0+8
2x = 8
2x/2 = 8/2
x=4
(Both factors would be x = 4; therefore, there is only one real solution. This parabola has a vertex on the
x-axis and that’s why there’s only one x-intercept.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 10: Quadratic Equations
2.
3x2 – x = -5
This equation must be set equal to 0 by adding 5 to both sides.
3x2 – x +5 = -5+5
3x2 –x +5 = 0
A=3
B = -1
C =5
The discriminant is: b2 – 4ac
(-1)2 – 4(3)(5) = -59 The discriminant is negative, therefore, the solutions are not
real numbers. Therefore, we will stop here. This parabola when graphed will not cross the x-axis.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com