# Math Help - solving for x intercept

1. ## solving for x intercept

to solve 7/(x-2) - 4/(x-1) + 3/(x+1) = 0
for the x intercepts, would I have to clear out all denominators and then use either factoring or quadratic formula to solve for x intercepts?

i used these steps to figure out the x intercepts but i keep getting the wrong answer with the back of the book AND when i graph the functions, they have the same x intercepts with the back of the book so i am guessing that it is right.

so my question is, are my steps wrong? because i did it several times to make sure there were no mechanical errors..

2. Originally Posted by skeske1234
to solve 7/(x-2) - 4/(x-1) + 3/(x+1) = 0
for the x intercepts, would I have to clear out all denominators and then use either factoring or quadratic formula to solve for x intercepts?

i used these steps to figure out the x intercepts but i keep getting the wrong answer with the back of the book AND when i graph the functions, they have the same x intercepts with the back of the book so i am guessing that it is right.

so my question is, are my steps wrong? because i did it several times to make sure there were no mechanical errors..
You need to multiply through by the least common denominator, which is $(x-2)(x+1)(x-1)$.

You should then end up with:

$7(x+1)(x-1)-4(x-2)(x+1)+3(x-2)(x-1)=0$ $\implies 7x^2-7-4x^2+4x+8+3x^2+3x-6=0$ $\implies 6x^2+7x-5=0\implies x=\frac{-7\pm13}{12}$

Thus, $x=\frac{1}{2}$ or $-\frac{5}{3}$

Does this make sense?

3. I do not believe I can over-emphasize the utility of knowing when to quit. When faced with a problem, you should ALWAYS keep in mind that the solution may be other than what you expect, or may not exist at all. The trick is simply PROVING that it doesn't exist.

Your trio of fractions should lead to $\frac{6x^{2}-5x+7}{stuff}$. With just a little effort, $(-5)^{2} - 4 \cdot 6 \cdot 7 = 25 - 168 = -143$, we are done. There are NO solutions for your expression set to zero (x-intercepts.)