A.5 Day 2 Solving Equations 2011
August 31, 2011
A.5 Solving Equations
Day 2
Objectives:
• Review solving by factoring.
• Solve quadratics by completing the square.
• Solve quadratics by using the quadratic formula.
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Review:
Solve the following equations.
1. 8x ­ (2x + 1) = 3x ­ 10
2. 4z3 ­ 8z2 = 0 3. x3 + 6x2 ­ 7x = 0
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Completing the Square
This is what you do when you are given a quadratic function in standard form and you want to write it in vertex form.
y = ax2 + bx + c
y = a(x ­ h)2 + k
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Example #1: Complete the square to find the vertex and the x­intercepts.
1) Isolate x2 and x on the left side of the equation.
2
x + 6x + 12 = 15
2
x + 6x = 3
2) Off to the side, compute: 2
( b2 )
b = 6, so...
2
( )
6
2
2
(3)
9
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A.5 Day 2 Solving Equations 2011
August 31, 2011
3) Add this number to both sides of the equation.
x 2 + 6x + 9 = 3 + 9
*Remember, you can do whatever you want to an equation as long as you do the same thing to both sides!
b
4) Factor on the left using 2
and simplify on the right.
x 2 + 6x + 9 = 12
b
(x + )
2
2
2
*If the term is positive, add it, if it's negative, subtract it.
(x + 3) = 12
2
5) Move the constant back to the left side of the equation.
(x + 3) ­ 12 = 0
6) Find the vertex.
(x + 3) ­ 12 = 0
2
Vertex: (­3, ­12)
7) Now solve to find the x­intercepts.
2
(x + 3) = 12
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A.5 Day 2 Solving Equations 2011
August 31, 2011
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A.5 Day 2 Solving Equations 2011
Example #2:
August 31, 2011
2
x + 3x ­ 8 = 0
b = 3
2
x + 3x = 8
3
2
2
( )
9
4
2
x + 3x + = 8 + 9
4
Keep this a fraction
= 9
4
Make this a decimal
9
4
2
x + 3x + = 8 + 2.25
2
3
(x+ ) = 10.25
2
2
3
(x+ ) ­ 10.25 = 0
2
Vertex: ( ­
3
2
, ­10.25 )
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A.5 Day 2 Solving Equations 2011
Example #3:
August 31, 2011
2
x ­ 6x + 12 = 9
Vertex: (3, ­6)
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A.5 Day 2 Solving Equations 2011
Example #4:
August 31, 2011
2
3x ­ 18x + 8= 0
2
3(x ­ 6x ) + 8= 0
Woa!!! a = 3
That's tricky!!
2
3(x ­ 6x ) = ­8
2
3(x ­ 6x + 9 ) = ­8 + 27
2
3(x ­ 3) = 19
2
3(x ­ 3) ­19 = 0
Vertex: (3, ­19)
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A.5 Day 2 Solving Equations 2011
Example #5:
August 31, 2011
2
4x ­ 16x + 9= 0
Vertex: (2, ­7)
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A.5 Day 2 Solving Equations 2011
Find the x­intercepts:
August 31, 2011
2
4x ­ 16x + 9= 0
2
4(x ­ 2) ­ 7= 0
2
4(x ­ 2) =7
2
(x ­ 2) = 7/4
(x ­ 2) = 7/4
+
x = 2 7/4
­
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Solve the equation by completing the square.
ax2 + bx + c = 0
x2 + b/ax + c/a = 0
x2 + b/ax = ­c/a
2
x2 + b/ax + b /4a 2 = ­c/a + b /4a 2
b
2
(x + /2a) = 2
­4ac
b 2
/4a 2 + /4a 2
2
(x + b/2a)2 = (­4ac +b )/4a 2
b
x + /2a = ± (
2
(b ­ 4ac)
/4a 2)
2
x = ­b/2a ± (b ­ 4ac)/2a
x = ­b ± b2 ­ 4ac
2a
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A.5 Day 2 Solving Equations 2011
August 31, 2011
The Quadratic Formula
Another way to find the solutions of a quadratic is to use the quadratic formula:
The quantity is called the discriminant of the quadratic equation.
1. If there are two unequal real solutions.
2. If there is a repeated real solution.
3. If there is no real solution.
http://www.youtube.com/watch?v=TBBBHel5Kdk&feature=related
http://www.youtube.com/watch?v=HgD9MWU5PBo&feature=related
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Example #6:
Find the real solutions, if any, of the equation. Since
3x2 + 1 = 5x
there are two real solutions.
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Example #7:
Use the discriminant to determine whether each quadratic has two unequal real solutions, a repeated real solution (exactly one solution), or no real solution without solving the equation.
A.
3x2 + 8 = 2x
B.
y = 2x2 + 5x ­ 1
C.
x2 + 2x + 1 = 0
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A.5 Day 2 Solving Equations 2011
August 31, 2011
Homework: page 999 (114 ­ 116, 119 ­ 121, 123 ­ 126, 145, 146)
&
A.5 Completing the Square Practice 1/2 sheet
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