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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 05:59
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What are the solutions to the following equation?

(x^2 + x - 12)/(x - 2) = 0

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I thought that I had to get rid of the denominator so I multiplied the equation with x-2 but appearantly that was wrong. Does anyone know how to solve this? :shock:

Thank youuu! :oops:

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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 06:08
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If you cross multiply, the numerator will be zero.

Numerator = x^2 + x - 12 = (x+4)(x-3) = 0

hence x = -4 or x = 3
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 06:15
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You can cross multiply here.

So we have x^2+x-12 = 0
=> x = 3 or -4
and x cannot be 2 because of x-2 in the denominator.

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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 11:42
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I think you were on the right track:

\(\frac{x^2+x-12}{x-2}=0\)

Multiply both sides by (x - 2):

\((x-2)\frac{x^2+x-12}{x-2}=0*(x-2)\)

The (x - 2) on the left cancels and the (x - 2) on the right is just 0. We're left with:

\((x^2+x-12) = 0\)

Now, you can simply the equation on the left into:

\((x-3)(x+4)=0\)

Finally, recall from algebra, you can separate this equation into 2 separate equations:
\(x-3=0\) and \(x+4=0\)

Just solve for each of the x's in the equations, and you are left with \(x = 3\), and \(x = -4\).

If you're curious why there are 2 x's, it's because the equation \((x^2+x-12)\) is the equation of a parabola, and the value at the points where x = 3 and x = -4, the y value of the point is 0.
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 14:46
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You can get rid of the bottom (x-2) but it's important to remember that the fraction WILL NOT work if you have x = 2, because any numerator divided by 0 is UNDEFINED

\(\frac{(x^2 + x - 12)}{(x-2)} = 0\)

Factor the top portion FIRST [you may be able to cross out terms]:

\(\frac{(x+4)(x-3)}{(x-2)} = 0\)

In this case, you actually can't

Multiply (x-2) on both sides then you get

(x+4)(x-3) = 0

x = -4, 3

So, we know that X = -4 and 3. Additionally, we know X DOES NOT equal -2
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 21:35
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To show steps leading to (x+4)(x-3)
=>x^2 + x - 12=0
=>x^2 + 4x -3x - 12=0
=>x(x + 4) -3(x + 4)=0
=>(x+4)(x-3) =0
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 22:00
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There are 3 values for x.

x^2+x-12 /x-2 = 0

First, looking at the equation. We see that x IS NOT equal to 2, because of x-2 as the denominator.

(x-4)(x+3)=0*(x-2)
(x-4)(x+3)=0

x = 4, x=-3, x<>2

I'm getting three values of x.
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 23:46
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lawschoolsearcher
There are 3 values for x.

x^2+x-12 /x-2 = 0

First, looking at the equation. We see that x IS NOT equal to 2, because of x-2 as the denominator.



(x-4)(x+3)=0*(x-2)
(x-4)(x+3)=0

x = 4, x=-3, x<>2

I'm getting three values of x.
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Well, I think if you used any number greater than two or less than two (other than 4,-3) for x the answer wouldn't be equal to 0.

For example

X=5, you get 8/3=0

And if you chose -5, you would get 18/7=0

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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 05 Apr 2011, 23:52
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X=-5, then -18/7. Silly mobile version won't let me edit

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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 06 Apr 2011, 02:07
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Thank you guys! I get it now.. I multiplied the upper part with x-2 aswell.. Don't ask me why I did that, I thought since there were parentheses around the (x-2) I couldnt cancel out before multiplying.. Hmmm
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(x^2 + x - 12)/(x - 2) = 0, What are the solutions for x? 17 Jul 2014, 01:55
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lawschoolsearcher
There are 3 values for x.

x^2+x-12 /x-2 = 0

First, looking at the equation. We see that x IS NOT equal to 2, because of x-2 as the denominator.

(x-4)(x+3)=0*(x-2)
(x-4)(x+3)=0

x = 4, x=-3, x<>2

I'm getting three values of x.
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The equation formed is incorrect

Values of x = -4 & 3 as computed by others above

3 values of x are NOT possible for this equation

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