5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
5.5 (Day 2) Quadratic Equations & 5.6 Complex Numbers
Objectives:
*Solve quadratic equations by finding square roots.
*Identify complex numbers.
*Add, subtract, and multiply complex numbers.
Nov 30­4:19 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Nov 30­7:57 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
So far we have been solving by factoring. Here's something new!
You can solve an equation in the form ax2 = c by finding square roots.
Example #1: Solve 5x2 ­ 180 = 0 by finding square roots. 5x2 ­ 180 = 0
5x2 = 180 x2 = 36
x = ±6
Nov 30­5:05 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #2: Solve each equation by finding square roots. a. 4x2 ­ 25 = 0
b. 3x2 = 24
c. x2 ­ 1/4 = 0
Nov 30­5:05 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Some quadratic equations have solutions that are complex numbers.
Example #3: Solve 4x2 + 100 = 0 by finding complex solutions. 4x2 + 100 = 0
4x2 = ­100 x2 = ­25
x = ±√­25
What should we do now?? We are trying to take the square root of a negative number!
x = ±5i
Nov 30­5:08 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
The imaginary number i is defined as the number whose square is -1.
i2 = -1 so i = √-1
An imaginary number is any number of the form a + bi,
where a and b are real numbers, and b ± 0.
Look at the new version...
Nov 30­5:10 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #4: Simplify √­8 by using the imaginary number i. √­8 = i(√8)
= i(2√2)
= 2i√2
Nov 30­5:11 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #5: Simplify each number by using the imaginary number i. a. √­2
b. √­12
c. √­36
Nov 30­5:11 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Imaginary numbers and real numbers make up the set of complex numbers.
Example #6: Write the complex numbers in the form a + bi.
a. √­9 + 6 b. √­18 + 7
3i + 6
3i√2 + 7
6 + 3i
7 + 3i√2 Nov 30­5:12 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
You can apply the operations of real numbers to complex numbers.
If the sum of two complex numbers is 0,
then each number is the opposite, or additive inverse, of the other.
Example #7: Find the additive inverse of ­2 + 5i.
a. ­2 + 5i
b. ­5i
c. 4 ­ 3i
­(­2 + 5i)
5i
­4 + 3i
2 ­ 5i
Nov 30­5:13 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
To add or subtract complex numbers, combine the real parts and the
imaginary parts separately.
Example #8: Simplify the expression.
a. (5 + 7i) + (­2 + 6i)
b. (8 + 3i) ­ (2 + 4i)
= 5 + 7i + (­2) + 6i
= 8 + 3i ­ 2 ­ 4i
= 3 + 13i
c. (4 ­ 6i) + 3i
= 6 ­ i
d. 7 ­ (3 ­ 2i)
= 4 ­ 6i + 3i
= 7 ­ 3 + 2i
= 4 ­ 3i
= 4 + 2i
Nov 30­5:39 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
For two imaginary numbers bi and ci, (bi)(ci) = bc(-1) = -bc.
You can multiply two complex numbers of the form a + bi
by using the procedure for multiplying binomials (FOIL).
Example #9: Multiply complex numbers.
a. (5i)(­4i) = ­20i2
b. (2 + 3i)(­3 + 5i)
= ­6 + 10i ­ 9i + 15i2
= ­20(­1)
= ­6 + i ­ 15
= 20
= ­21 + i
Nov 30­5:40 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #10: Simplify each expression.
a. (12i)(7i) b. (4 ­ 9i)(4 + 3i)
= 84i2
= 16 + 12i ­ 36i ­ 27i2
= 84(­1)
= 16 ­ 24i + 27
= ­84
= 43 ­ 24i
Nov 30­5:40 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #11: Solve by finding complex solutions. a. 3x2 + 48 = 0
b. ­5x2 ­ 150 = 0
c. 8x2 + 2 = 0
3x2 = ­48
­5x2 = 150
8x2 = ­2
x2 = ­16
x2 = ­30
x2 = ­2/8
x = √­16
x = √­30
x2 = ­1/4
x = 4i
x = i√30
x2 = √­1/4
x = 1/2 i
Nov 30­5:41 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
Example #12:
Nov 30­5:05 PM
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
December 03, 2008
HOMEWORK:
Practice 5-5
&
Practice 5-6
Nov 30­7:41 PM
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