Irrational Equation - I end up in a dead end

[Hivoyer]

Homework Statement

This is the equation:
2/(2 - x) + 6/(x^2 - x - 2) = 1

Homework Equations

sqrt= square root
^ = to the power of

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

 

[DrClaude]

2x^2 - 2x - 2 + 12 - 6x = (2 - 2)(x^2 -x -2)

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?

 

[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.

 

[DrClaude]

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

You've made an error yourself

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

 

[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.

 

[DrClaude]

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.

That is incorrect. Please show your work.

 

[Curious3141]

Homework Statement

This is the equation:
2/(2 - x) + 6/(x^2 - x - 2) = 1

Homework Equations

sqrt= square root
^ = to the power of

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)

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.

 

[Ray Vickson]

Homework Statement

This is the equation:
2/(2 - x) + 6/(x^2 - x - 2) = 1

Homework Equations

sqrt= square root
^ = to the power of

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

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