1. Jul 15, 2010

### Cycloned

Need help solving some. I'll put up one for now.

a/b-x + b/a-x = 2

I'm just looking for how to solve it, because I always reach

2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0

Thanks!

2. Jul 15, 2010

### l'Hôpital

That = 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0

Do you know the quadratic equation?

3. Jul 15, 2010

### Cycloned

I know quadratic equations, but I'm getting stuck here, because I do not know how to solve for x.

I did establish the fact that 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0
is correct because it follows ax^2 + bx + c.

4. Jul 15, 2010

### Staff: Mentor

What this means is probably different from what you intended. When you write fractions in a single line of text, use parentheses if the numerator or denominator have more than one term.

What you wrote would be interpreted as this:
$$\frac{a}{b - x} + \frac{b}{a - x} = 2$$
I get something different - different coefficients for the two terms in x, and different signs for the a^2 and b^2 terms. Check your work.

5. Jul 15, 2010

### Staff: Mentor

2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0 IS a quadratic equation. The question should be: Do you know the quadratic formula?

The quadratic formula gives solutions to the equation ax^2 + bx + c = 0. Since both the quadratic formula and the equation above use a and b, be careful in what you call a, b, and c in the quadratic formula.

As already noted, some of the coefficients in the OP's equation are incorrect.

Also, since the equation in the original post involved division by a - x and b - x, it should be stated explicitly that x can't be a, and x can't be b.