Download Solving Quadratic Equations By Formula Key

Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Real
part
Imaginary
part
a _
Solving Quadratic Equations by Formula
bi
KEY
Complex Numbers
All the quadratic equations solved to this point have had two real solutions or roots. In some cases,
solutions involved a “double root,” but there were always two solutions.
Study the quadratic equation
0 = x2 + x + 2.
•
Can this equation be factored in order to find the solutions or roots?
No, because the polynomial cannot be factored.
•
Can the zeros be found from the graph of the function y = x2 + x + 2? Explain.
No, because the graph of the function does not cross the x-axis.
There are still two solutions to this quadratic equation. However, they are not real numbers.
Instead, the solutions involve numbers that are imaginary.
The imaginary number unit is designated as “i.”
There are two ways that we define i.
Words
Symbols
•
“i” is the number you can square to get an answer of -1.
•
Or, “i” is the square root of -1
i 2 = -1
i=
−1
Square roots of negative numbers can be simplified using “i”:
− x = (−1) x = −1 × x = i x
• If x is a non-negative real number, then
When you have a negative number under
radical,
you can take the negative out as an “i.”
a
Example Problems: Simplify the following complex numbers by writing them in terms of i.
A)
­ 49
©2012, TESCCC
=
7i
Check: ( 7i )2 = 49i2 = 49(-1) = -49
09/06/12
page 1 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
B)
­ 121
C)
=
17 ­ 42
D)
­ 3
E)
11i
=
=
12 + ­ 36
2
©2012, TESCCC
Check: ( 11i)2 = 121i2 = 121(-1) = -121
5i
3
i
6+3i
09/06/12
page 2 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Formula KEY
A complex number is the sum of a real number and an
imaginary number, usually written in the form a + bi.
Write the following numbers in a + bi form, and identify the real and imaginary parts. Simplify first if
necessary.
F) 25
25 + 0i
Real part = 25
Imaginary part = 0i
9 + ­ 144
6
I)
1.5 + 2i
Real part = 1.5
Imaginary part = 2i
G) 25i
0 + 25i
Real part = 0
Imaginary part = 25i
H) 3 – 5i
3 – 5i
Real part = 3
Imaginary part = -5i
J) 3 + 5i – 7 + 8i
K) (6 + 2i)(6 – 2i)
-4 + 13i
Real part = -4
Imaginary part = 13i
40
Real part = 40
Imaginary part = 0i
The conjugate of a complex number has the same real
2 + 3i
part but an opposite imaginary part. (For example,
2 ­ 3i
is the complex conjugate of
.)
Complex Conjugates
a + bi
and
a ­ bi
* When complex conjugates are added—or multiplied—the result will have an imaginary part of 0 i.
Tell the conjugate of each complex number.
Number
Conjugate
Number
Conjugate
L) 25
25
N) -3 + 6i
-3 – 6i
M) 25i
-25i
O) -12 – 5i
-12 + 5i
• Solving Equations with Complex Numbers
Some quadratic equations can be solved by isolating the variable
and taking the square root of both sides.
©2012, TESCCC
09/06/12
P)
Q)
Number
2
­ 51 i
3
5+ i 3
2
3
Conjugate
+ 51 i
5 −i 3
x 2 = ­ 100
Sample:
x = ± ­ 100
x = ± 10i
page 3 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Why are the solutions complex conjugates?
6 x 2 + 96 = 0
R) Solve
x = 4i or x = -4i
©2012, TESCCC
Answers will vary as to explaining why they
are complex conjugates.
Sample: Have positive and negative
imaginary part.
09/06/12
page 4 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Formula KEY
Describe the complex numbers below by filling in the table.
Chart A
­ 121
1)
2) -56
3)
85 + ­ 169
­
­
16
49
4)
5)
­ 5
6) 3 – 14i – 9 + 5i
7) (5 + 2i) 2
Simplified
Real Part
Imaginary Part
Complex
Conjugate
11i
0
11i
-11i
-56
-56
0
-56
85 +13i
85
13i
85 – 13i
−
4
i
7
4
i
7
4
i
7
0
−
i 5
0
i 5
−i 5
-6 – 9i
-6
-9i
-6 + 9i
21 + 20i
21
20i
21 – 20i
Solve the following quadratic equations and describe the quadratic solutions.
8)
2 x 2 + 50 = 0
9)
5i
Complex conjugates
10)
6i
Complex conjugates
What is the quadratic equation with solutions of
(x – (7i)) (x – (-7i)) = 0
(x – 7i) (x + 7i) = 0
x2 + 7ix – 7ix – 49i2 = 0
x2 – 49i2 = 0
x2 + 49 = 0
©2012, TESCCC
­ 3 x 2 ­ 8 = 100
± 7i
04/18/13
?
page 5 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Formula KEY
Part III: Quadratic Formula
Any quadratic equation—including those with imaginary or complex solutions—can be solved with the
quadratic formula.
If ax2 + bx + c = 0, and a  0, then Steps:
• Get one side of the equation equal to zero.
Quadratic
2
• Identify the values of a, b, and c.
­ b ± b ­ 4ac
Formula
x=
• Substitute these values into the formula.
2a
• Simplify to obtain the two solutions.
b 2 ­ 4ac
From the quadratic formula,
is called the discriminant. Because this part of the formula lies
under the radical, it determines the type of solutions the quadratic equation will have.
If…
Then the equation has…
Sample
x + 6x + 9 = 0
a = 1, b = 6, c = 9
Previous Methods
x 2 + 6x + 9 = 0
2
b 2 ­ 4ac = 0
b 2 ­ 4ac > 0
One rational double root
,
b 2 ­ 4ac
and
is a
perfect square
(1, 4, 9, 16, 25, etc.)
b 2 ­ 4ac > 0
,
b 2 ­ 4ac
and
is a
non-square
(2, 5, 8, 12, etc.)
b 2 ­ 4ac < 0
Two rational roots
(6) ­ 4(1)(9 ) = 0
Two complex roots
(Conjugates involving
imaginary numbers)
x = -3 (double root)
2x 2 + x − 3 = 0
2x 2 + x ­ 3 = 0
a = 2, b = 1, c = ­ 3
(2 x + 3)( x − 1) = 0
3
x = − , x =1
2
(1) 2 ­ 4(2)( ­ 3) = 25
x 2 ­ 10 x + 23 = 0
Two irrational roots
( x + 3)( x + 3) = 0
2
a = 1, b = ­ 10, c = 23
•
•
Will not factor
Graphs can provide
approximations only
5± 2
(Here, x =
)
•
•
Will not factor
Graphs cannot
provide solutions
2 ± 6i
(Here, x =
)
( ­ 10) 2 ­ 4(1)( 23 ) = 8
x 2 ­ 4 x + 40 = 0
a = 1, b = ­ 4, c = 40
( ­ 4) 2 ­ 4(1)( 40) = ­ 144
Example Problems
For each problem,
a) Write the equation in the form ax2 + bx + c = 0.
b) Find the discriminant, and use it to describe the nature of the roots of the equation.
c) Complete the quadratic formula to find the solutions.
©2012, TESCCC
04/18/13
page 6 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
1)
3x 2 + 7x = ­ 2
©2012, TESCCC
a) 3x2 + 7x + 2 = 0
b) Discriminant =25 (Two rational roots)
1
−
3
c) Roots: {-2,
}
04/18/13
page 7 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Formula KEY
2)
x 2 ­ 2x = 2x ­ 2
a) x2 – 4x + 2 = 0
b) Discriminant = 8 (Two irrational roots)
4± 8
2
c) Roots: x =
, or approximately {3.4, 0.6}
3)
x 2 = 14 x ­ 58
a) x2 – 14x + 58 = 0
b) Discriminant = -36 (Two complex conjugate roots)
c) Roots: {7 + 3i, 7 – 3i}
4)
Given a quadratic equation with rational coefficients and a root of 3 + 5i
a) What is another root to the equation?
b) What are the factors of the equation?
c) What is the quadratic equation with the given root?
d) What is the representative function?
a) 3 – 5i
b) (x – (3 + 5i))(x – (3 – 5i))
c) x2 – 6x + 34 = 0
d) y = x2 – 6x + 34
5)
Given a quadratic equation with rational coefficients and a root of -2 + 8i
a) What is another root to the equation?
b) What are the factors of the equation?
c) What is the quadratic equation with the given root?
d) What is the representative function?
a) -2 – 8i
b) (x – (-2 + 8i))(x – (-2 – 8i))
c) x2 + 4x + 68 = 0
©2012, TESCCC
04/18/13
page 8 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
d) y = x2 + 4x + 68
©2012, TESCCC
04/18/13
page 9 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Formula KEY
Practice Problems
For each problem,
 Use the discriminant to describe the nature of the roots of the equation.
 Complete the quadratic formula to find the solutions.
1)
x 2 − 8 x + 12 = 0
2)
Discriminant > 0, perfect square
Two rational roots: {6, 2}
3 x 2 + x = 10
Discriminant > 0, perfect square
5
3
Two rational roots: {-2,
3)
x 2 + 25 = 10 x
4)
Discriminant = 0
One rational double root: {5}
5)
x 2 + 6 = 8x
10 x 2 + 2 x + 5 = 0
Discriminant < 0
Two complex conjugate roots:
−1 ± 7i
10
{
} or {0.1 + 0.7i, 0.1 – 0.7i}
6)
Discriminant > 0, non-square
4 ± 10
Two irrational roots: {
}
2 x 2 + x − 14 = −11
Discriminant > 0, perfect square
3
2
Two rational roots: {- , 1}
7)
Given a quadratic equation with rational coefficients and a root of
2
3i
a) What is another root to the equation?
( x − 32 i )( x + 32 i )
b) What are the factors of the equation?
x 2 + 94 = 0
c) What is the quadratic equation with the given root?
©2012, TESCCC
}
04/18/13
2
− i
3
page 10 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
8)
Given a quadratic equation with rational coefficients and a root of 5 – i
a) What is another root to the equation? 5 + i
b) What are the factors of the equation? (x – (5 – i)) (x – (5 + i))
c) What is the quadratic equation with the given root? x2 – 10x + 26 = 0
©2012, TESCCC
04/18/13
page 11 of 11