5.5 (day 2) quadratic equations & 5.6 complex...
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5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
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December 03, 2008
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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.
5.5 Day 2 Solving Quadratics & 5.6 Complex Numbers
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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
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Example #2: Solve each equation by finding square roots.
a. 4x2 25 = 0 b. 3x2 = 24 c. x2 1/4 = 0
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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
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Look at the new version...
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.
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Example #4: Simplify √8 by using the imaginary number i.
√8 = i(√8)
= i(2√2)
= 2i√2
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Example #5: Simplify each number by using the imaginary number i.
a. √2 b. √12 c. √36
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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
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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
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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 = 6 i
c. (4 6i) + 3i d. 7 (3 2i)
= 4 6i + 3i = 7 3 + 2i
= 4 3i = 4 + 2i
To add or subtract complex numbers, combine the real parts and the imaginary parts separately.
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For two imaginary numbers bi and ci, (bi)(ci) = bc(-1) = -bc.
You can multiply two complex numbers of the form a + biby using the procedure for multiplying binomials (FOIL).
Example #9: Multiply complex numbers.
a. (5i)(4i) b. (2 + 3i)(3 + 5i)
= 20i2 = 6 + 10i 9i + 15i2
= 20(1) = 6 + i 15
= 20 = 21 + i
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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
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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
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Example #12:
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HOMEWORK:Practice 5-5
&Practice 5-6
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