View Full Version : Find the set of values for which the following expression is

LMande

07-07-2006, 10:25 PM

Here's the question...

Find the set of values for which the following expression is defined.

x^2 - 9 / x^2 + 2x - 63

I believe here I'm supposed to set the denominator to equal to zero.

So I go...

x^2 +2x - 63 = 0 Then solve for x.

So I add 63 to both sides...

x^2 +2x = 63

I factor..

(x+2x)(x+1) =63

But now I'm stumped....

I know this must seem so easy to y'all but I've been out of school too long! LOL

Any help on this would be appreciated.

stapel

07-07-2006, 11:37 PM

x^2 - 9 / x^2 + 2x - 63

The above means this:

. . . . .\L x^2\,-\,\frac{9}{x^2}\, +\,2x\,-\,63

Is this what you meant? Or did you mean something like the following:

. . . . .\L \frac{x^2\,-\,9}{x^2\,+\,2x\,-\,63}

...or something else?

Thank you.

Eliz.

Mrspi

07-08-2006, 09:26 AM

Here's the question...

Find the set of values for which the following expression is defined.

x^2 - 9 / x^2 + 2x - 63

I believe here I'm supposed to set the denominator to equal to zero.

So I go...

x^2 +2x - 63 = 0 Then solve for x........YEP......but from here on, you are WAY off base

So I add 63 to both sides...

x^2 +2x = 63

I factor..

(x+2x)(x+1) =63

But now I'm stumped....

I know this must seem so easy to y'all but I've been out of school too long! LOL

Any help on this would be appreciated.

Did you notice that your equation is a quadratic? You WANT one side to be equal to 0:

x<sup>2</sup> + 2x - 63 = 0

Factor the left side.....and based on previous posts, it is obvious that you need to do MUCH review and practice on factoring!

(x - 7)(x + 9) = 0

Set each factor equal to 0 and solve for x. The values you get for x will be the values of x for which the rational expression is undefined; it is defined for all other real numbers except those two.

LMande

07-08-2006, 10:36 AM

I must agree! I need lots of review on factoring! LOL I truly appreciate all your help though! It's like when I screw up then read your responses I go.. "well duh!" I guess I'm just one that has to learn by making the mistakes.

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