# Math Help - Need help with an equation based on quadratic formula

1. ## Need help with an equation based on quadratic formula

I have the following problem and I've been trying to solve it

$\frac{6}{r^2-1}-\frac{1}{2}=\frac{1}{r+1}$

but when I try solving I just can't figure out how they get to 3 and -5. I'm supposed to solve it turning it into

$ax^2+bx+c=0$

I need to get rid of these denominators and therefore multiply the whole expression by its least common denominator

$2(r^2-1)(r+1)$

but I stay stucked...!! Need guidance.

2. you're probably confusing signs... try to take 1/r+1 to the ledtside and 1/2 to the right, the factorise 1/r+1 and derive the quadratic equation ...what do you get?

3. Note that $r^2-1 = (r-1)(r+1)$ (difference of two squares)

This means that the denominator is $r^2-1$

$\dfrac{6}{r^2-1} - \dfrac{1}{2} \cdot \dfrac{(r^2-1)}{r^2-1} = \dfrac{r-1}{r^2-1}$

Then multiply by 2 to get $\dfrac{12}{r^2-1} - \dfrac{(r^2-1)}{r^2-1} = \dfrac{2(r-1)}{r^2-1}$

The denominator will cancel and then should be easy to simplify and solve - just remember that $r \neq \pm 1$ as this would entail division by 0

4. ## Haven't found my way yet

I did what you said but cannot find the x = 3 , -5 yet. Can anyone please tell me where my mistake is?

5. Originally Posted by Alienis Back
$\frac{6}{r^2-1}-\frac{1}{2}=\frac{1}{r+1}$
Start this way (easier LCM):
6 / (r^2 - 1) = 1 / (r + 1) + 1/2 ; combine right side:
6 / (r^2 - 1) = (r + 3) / [2(r + 1)] ; factor left side:
6 / [(r + 1)(r - 1)] = (r + 3) / [2(r + 1)] ; cancel out the (r + 1)'s:
6 / (r - 1) = (r + 3) / 2 ; crossmultiply:
(r - 1)(r + 3) = 12 ; expand left side:
r^2 + 2r - 3 = 12

Can you wrap it up? I'm sure you can.