## Solving Rational Equations (Imaginary Answer?)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Gsingh
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Joined: Wed Aug 08, 2012 5:48 pm
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### Solving Rational Equations (Imaginary Answer?)

Problem:
$\frac{x}{x\, -\, 2}\, +\, \frac{2x}{4\, -\, x^2}\, =\, \frac{5}{x\, +\, 2}$

I can usually solve these kinds of problems pretty easily, but I got stuck with this one and when I checked the answer it was imaginary...

I made a common denominator of -(x-2)(x+2) on the left side and then combined to form one fraction, then i cross multiplied and somehow i ended up with x^2(x-3) = -20 and don't know what to do now.

The answer is supposed to be $\left(\frac{5}{2}\right)\,\pm\, \frac{i\, \sqrt{15}}{2}$

Thanks for the help

stapel_eliz
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I made a common denominator of -(x-2)(x+2) on the left side and then combined to form one fraction, then i cross multiplied and somehow i ended up with x^2(x-3) = -20

Gsingh
Posts: 5
Joined: Wed Aug 08, 2012 5:48 pm
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### Re: Solving Rational Equations (Imaginary Answer?)

Ok these were my steps... (im going to explain in words instead of pictures because it took me forever to post that last message)

1. factored the 4-x^2 into -(x+2)(x-2).
2. then i multiplied top and bottom of x/(x-2) by -(x+2) so now it also has the -(x+2)(x-2) denominator.
3. then i combined the fractions to get -x^2-2x+2x/-(x+2)(x-2) on the left which is still equal to 5/(x+2) on the right.
4. i simplified to get -x^2/-(x^2-4) = 5/(x+2)
5. cross multiply... -x^2(x+2) = -5(x^2-4)
6. which simplifies to.. x^3 + 2x^2 = 5x^2 - 20 (i divided out the negative)
7. x^3 - 3x^2 = -20
8. x^2(x-3) = -20

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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Ok these were my steps... (im going to explain in words instead of pictures because it took me forever to post that last message)

1. factored the 4-x^2 into -(x+2)(x-2).
Simpler: -(x - 2) = 2 - x, and x - 2 = -(2 - x), so convert the original equation to:

. . . . .$\frac{x}{x\, -\, 2}\, -\, \frac{2x}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{5}{x\, +\, 2}$
4. i simplified to get -x^2/-(x^2-4) = 5/(x+2)
5. cross multiply... -x^2(x+2) = -5(x^2-4)
You might want to check the cancellation on the right-hand side. One of the factors of x2 - 4 should have cancelled off!

. . . . .$\frac{x}{x\, -\, 2}\, -\, \frac{2x}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{5}{x\, +\, 2}$

. . . . .$\frac{x(x\, +\, 2)}{(x\, -\, 2)(x\, +\, 2)}\, -\, \frac{2x}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{5(x\, -\, 2)}{(x\, -\, 2)(x\, +\, 2)}$

Now try multiplying through.

Gsingh
Posts: 5
Joined: Wed Aug 08, 2012 5:48 pm
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### Re: Solving Rational Equations (Imaginary Answer?)

Oh yeah, I didn't see the (x+2) that would cancel on both sides...thanks for the help!