I can't find where I'm making my mistake.

C

Chocolate

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Sep 2, 2005
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Solve [ (2x^2) / (x^2 - 9) ] - [ x / (3 - x) ] = -5 / (x + 3)

This is what I've tried:

NPV= +/- 3
[ (2x^2) / (x + 3)(x - 3) ] - [x / (3 - x) ] = -5 / (x + 3)

I multiplied this by it's LCF to get:

2x^2(3 - x) - x(x + 3)(x - 3) = -5(x - 3)(3 - x)

(6x^2 - 2x^3) - (x^2 + 3x)(x-3) = (-5x + 15)(3 - x)

(6x^2 - 2x^3) - (x^3 - 3x^2 + 3x^2 - 9x) = -15x + 5x^2 + 45 - 15x

(6x^2 - 2x^3) - (x^3 - 9x) = 5x^2 - 30x + 45

6x^2 - 2x^3 - x^3 + 9x = 5x^2 - 30x + 45

-3x^3 + 6x^2 + 9x - 5x^2 + 30x - 45 = 0

-3x^3 + x^2 + 39x - 45 = 0

P(3) = -3(3)^3 + (3)^2 + 39(3) - 45

P(3) = 0

3 | -3 1 39 -45
| 9 -24 45
-3 -8 15 0
 
The LCF is (x+3)(x-3) but you didn't cancel the x-3 in the first term denominator. You kept x-3 in the numerator.
2x^2 + x(x + 3) = -5(x - 3)
 
I thought the LCF was (x - 3)(x + 3)(3 - x) that's why i only canceled (x - 3)(x + 3)
 
Nope it is the highest power of each factor. By your reasoning it would be (x+3)(x-3)(x+3)(x-3) but you only need one of each 'cause that's all there is in any denominator.
 
What about (3 - x), is it the same as (x - 3) but then the answer would be x = -5 and that's incorrect because the ans key was (-4 +/- sqroot 61) / -3
 
Ignore my equation in that first reply. I was trying to do it in edit and got mixed up.
You change 3-x to x-3 and change the sign of the x

[(2x^2)/(x^2-9)]-[x/(3-x)] = -5/(x+3)
[(2x^2)/(x^2-9)]+[x/(x-3)] = -5/(x+3)
2x^2 + x(x+3) = -5(x-3)
 
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