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September 8th, 2011, 06:13 PM   #1
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Completing the square

I'm pretty much at my wit's end... I'm taking college algebra and definitely recognize this but just can't remember a good method. I took remedial algebra last year and am finally in regular algebra, but have forgotten parts of it xD

Anyways, I'm sure you all have these questions ALL the time but I'd really appreciate it if someone took the time to explain this to me. I'll be going to the free tutoring ASAP and as soon as it opens next week, but this is due on Sunday... between this and friends I'm hoping to figure it out.

Here it is:
8x^2 + 6x = 9

I've tried a few different methods and nothing seems to work--it turns into this massive fraction that I can't make sense of. I'm good up until the step where it says "divide by two and then square it to complete the square." Sounds easy enough but I cannot for the life of me figure out that in fraction form.

Any help greatly appreciated! Thanks!
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September 8th, 2011, 06:33 PM   #2
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Re: Completing the square

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September 8th, 2011, 06:39 PM   #3
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Re: Completing the square

8x+6x-9 = 0
Divide by 2:
4x+3x-4 = 0

Now pretend 4x+3x is a square (px+q) with the constant missing.
(px+q) = (p)x + (2pq)x + q = 4x+3x+??
Equating coefficients, p=4 so p=2, and 2pq=3 so q=3/4.
Then q=9/16 is the missing constant.

4x+3x
= 4x+3x +9/16 -9/16
= (2x+3/4) - 9/16

Therefore, substituting into the original equation:
4x+3x - 4 = 0
(2x+3/4) - 9/16 - 4 = 0
(2x+3/4) = 9/16 + 4 = 5 + 1/16

Taking the square root of both sides:
2x+3/4 = ?(81/16) = 9/4
2x = (-39)/4
x = (-39)/8 = -1 or 3/4
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