Completing the square
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Transcript of Completing the square
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Completing the Square
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EXITBACK
The easiest quadratic equations to solve are of the type
2 2x rwhere r is any constant.
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The solution to is
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x r
The answer comes from the fact that we solve the equation by taking the square root of both sides
2 2x r
2 2x r
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EXITBACK
Since is a constant, we get
where both sides are positive, but since is a variable it could be negative, yet is positive and is also positive.
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2r rx
2x
r
2x
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If we write and is negative, we are saying that a positive number is negative, which it cannot be.
To get around this contradiction we need to insist that
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2x x x
2x x x
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satisfies all possibilities since if is positive we use
and if is negative we use
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2x xx
2x xx
2x x
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EXITBACK
Thus, we get the solution for
We generally change this equation by multiplying both sides by , then we simplify and get
2 2x r
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2 2x rx r
1
x r
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We generally skip all the intermediate steps for an equation like
and get
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2 16x 4x
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EXITBACK
For quadratic equations such as
we solve and get
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29 25x
3 5x 5
3x
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EXITBACK
For quadratic equations that are not expressed as an equation between two squares, we can always express them as
If this equation can be factored, then it can generally be solved easily.
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2 0ax bx c
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If the equation can be put in the form
then we can use the square root method described previously to solve it. The solution for this equation is
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2 2( )k x m n
The sign of m needs to be the
opposite of the sign used in ( )x m
nx m
k
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The question becomes: “Can we change the equation from the formto the form ?”
Fortunately the answer is yes!
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2 0ax bx c 2 2( )x m n
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The procedure for changingis as follows. First, divide by , this gives
Then subtract from both sides. This gives
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2 0ax bx c a
2 0b c
x xa a
c
a
2 b cx x
a a
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We pause at this point to review the process of squaring a binomial. We will use this procedure to help us complete the square.
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EXITBACK
Recall that
If we let
we can solve for to get
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2 2 2( ) 2x r x rx r
2b
ra
r
2
br
a
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EXITBACK
Substituting in we get
Using the symmetric property of equations to reverse this equation we get
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2
br
a 2 2 2( ) 2x r x rx r
22 2
2( )
2 4
b b bx x x
a a a
22 2
2( )
4 2
b b bx x x
a a a
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EXITBACK
Now we will return to where we left our original equation. If we add to both sides ofwe get
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2
24
b
a2 b cx x
a a
2 22
2 24 4
b b c bx x
a a a a
2
2
4
4
b ac
a
22
2
4( )
2 4
b b acx
a a
or
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EXITBACK
We can now solve this by taking the square root of both sides to get
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2 4
2 2
b b acx
a a
2 4
2 2
b b acx
a a
2 4
2
b b acx
a
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EXITBACK
is known as the quadratic formula. It is used to solve any quadratic equation in one variable.
We will show how the quadratic equation is used in the example that follows.
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2 4
2
b b acx
a
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EXITBACK
First, we start with an equation
Then we change it to
From this we get
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23 11 20x x
23 11 20 0x x
3, 11, 20a b c
Remember 2 0ax bx c
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We then substitute into the quadratic formula, simplify and get our values for .
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3, 11, 20a b c
x
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Remember the quadratic equation is
and the values area = 3, b = 11, c = -20
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Doing so we get
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211 11 4(3)( 20)
2(3)x
11 121 240
6
11 361
6
11 19
6
2 4
2
b b acx
a
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EXITBACK
This gives us two values for ,
and
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x
11 19 4
6 3x
11 195
6x
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EXITBACK
The equation can be factored into which will give us the same solutions as the quadratic formula.
However, the beauty of using the quadratic formula is that it works for ALL quadratic equations, even those not factorable (and even when is negative).
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23 11 20x x (3 4)( 5) 0x x
2 4b ac
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EXITBACK
To review—the steps in using the quadratic formula are as follows:
1. Set the equation equal to zero, being careful not to make an error in signs.
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23 11 20x x 23 11 20 0x x
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2. Determine the values of a, b, and c after the equation is set to zero.
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23 11 20 0x x
3, 11, 20a b c 2 0ax bx c Remember
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3. Substitute the values of a, b, and c into the quadratic formula.
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211 11 4(3)( 20)
2(3)x
3, 11, 20a b c 2 4
2
b b acx
a
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4. Simplify the formula after substituting and find the solutions.
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211 11 4(3)( 20)
2(3)x
11 121 240
6
11 361
6
11 19
6
11 19 4
6 3x
11 195
6x