# Irrational Equation - I end up in a dead end

1. May 29, 2013

### Hivoyer

1. The problem statement, all variables and given/known data
This is the equation:
2/(2 - x) + 6/(x^2 - x - 2) = 1

2. Relevant equations
sqrt= square root
^ = to the power of

3. The attempt at a solution
First thing that comes to mind is to turn it into this:
2x^2 - 2x - 2 + 12 - 6x = (2 - 2)(x^2 -x -2)

Then it gets real ugly:
x^3 - x^2 - 8x + 14 = 0

If only I could somehow get rid of that +14, I would then divide it all by x and get a normal quadric equation...
x(x^2 - x - 8) + 14 = 0

Last edited: May 29, 2013
2. May 29, 2013

### Staff: Mentor

That should be
$$2x^2 - 2x -4 +12 -6x = (2-x)(x^2-x-2)$$

But have you tried dividing $x^2-x-2$ by $2-x$ first?

3. May 29, 2013

### AGNuke

$$\frac{2}{2-x}+\frac{6}{x^2-x-2}=1$$
$$\frac{2x^2-2x-4+ 12-6x}{(2-x)(x^2-x-2)}=1$$

You made an error in the first step itself. That's why you are stuck. Try correcting it.

Last edited: May 29, 2013
4. May 29, 2013

### Staff: Mentor

$$\frac{2x^2-2x-4+ 12-6x}{(2-x)(x^2-x-2)}=1$$

5. May 29, 2013

### Hivoyer

thanks for the help, I eventually end up with
(2x-4)/(x^2 - x - 2) = 1

which I turn into
x^2 - x + 2 = 0
and turns out that D < 0, so no real roots.

6. May 29, 2013

7. Jun 4, 2013

### Curious3141

Don't expand the LHS. Factorise. Same for the RHS.

After factoring both sides, you can group terms on the LHS.

You'll find that $(2-x)^2$ is common on both sides.

8. Jun 4, 2013

### Ray Vickson

In the original form of the problem, simplify the left-hand-side as much as possible before doing anything else.