Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 8.4 Quadratic Formula.
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Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 8.4 Quadratic Formula
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Transcript of Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 8.4 Quadratic Formula.
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Section 8.4
Quadratic Formula
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Objectives
• Solving Quadratic Equations
• The Discriminant
• Quadratic Equations Having Complex Solutions
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
, 0 and 1,xf x a a a
The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by
QUADRATIC FORMULA
2 4.
2
b b acx
a
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically.SolutionSymbolic SolutionLet a = 4, b = 3 and c = − 8.
2 4
2
b b acx
a
23 3 4 4 8
2 4x
3 137
8x
3 137
8x
or 3 137
8x
1.1x 1.8x or
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example (cont)
4x2 + 3x – 8 = 0Graphical Solution
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically.SolutionLet a = 3, b = −6 and c = 3.
2 4
2
b b acx
a
26 6 4 3 3
2 3x
6 0
6x
1x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically.SolutionLet a = 2, b = 4 and c = 5.
2 4
2
b b acx
a
24 4 4 2 5
2 2x
4 24
4x
There are no real solutions
for this equation because
is not a real number.
24
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
, 0 and 1,xf x a a a
To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.
THE DISCRIMINANT AND QUADRATIC
EQUATIONS
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula.Solution−2x2 + 5x − 3 = 0Let a = −2, b = 5 and c = −3.
Thus, there are two solutions.
b2 – 4ac
= (5)2 – 4(−2)(−3) = 1
2 4
2
b b acx
a
5 1
2 2x
4
4x
1x
or
6
4x
1.5x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
, 0 and 1,xf x a a a
If k > 0, the solution to x2 + k = 0 are given by
THE EQUATION x2 + k = 0
.x i k
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve x2 + 17 = 0.
SolutionThe solutions are
17.i
17 or 17.x i i
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. SolutionLet a = 3, b = −7 and c = 5. 2 4
2
b b acx
a
7 11
6x
and
27 7 4 3 5
2 3x
7 11
6
ix
7 11
6 6x i
7 11
6 6x i
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve Write your answer in standard form: a + bi. SolutionBegin by adding 2x to each side of the equation and then multiply by 5 to clear fractions.
Let a = −2, b = 10 and c = −15.
223 2 .
5 x
x
22 10 15 0x x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example (cont)
Let a = −2, b = 10 and c = −15.
223 2 .
5 x
x
2 4
2
b b acx
a
10 20
4x
210 10 4 2 15
2 2x
10 2 5
4
ix
5 5
2 2x i
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Solve by completing the square.SolutionAfter applying the distributive property, the equation becomes
Since b = −4 ,add to each side of the equation.
4 5x x
2 4 5.x x
242 4
2 4 4 5 4x x 2
2 1x 2 1x 2x i
2x i
The solutions are 2 + i and 2 − i.
