How to solve this quadratic equation?

Question

So I've got this quadratic equation and am totally unable to solve it. Can someone tell me how to do it?

$$\frac{a}{ax-1} + \frac{b}{bx-1} = a + b,$$

where $x$ is not equal to $\frac{1}{a}$ or $\frac{1}{b}$. We need to solve for $x$.

Answer 1

$$\frac{a}{a x -1}-b =a-\frac{b}{b x-1}$$

$$\frac{a b x-a-b}{a x-1} +\frac{a b x-a - b}{b x-1}=0$$

$$(a b x-a-b)\left(\frac{1}{a x-1} +\frac{1}{b x-1}\right)=0$$

$$(a b x-a-b)((a+b)x-2 )=0$$

$$x=\left\{\frac{2}{a+b},\frac{a+b}{a b}\right\}$$

Answer 2

$$\frac{a}{ax-1}+\frac{b}{bx-1}{=a+b} \Rightarrow \frac{a(bx-1)+b(ax-1)}{(ax-1)(bx-1)}=a+b \Rightarrow \frac{abx-a+abx-b}{(ax-1)(bx-1)}=a+b \Rightarrow \frac{2abx-a-b}{(ax-1)(bx-1)}=a+b \Rightarrow 2abx-a-b=(ax-1)(bx-1)(a+b)$$

Can you continue?

Answer 3

if you multiply the equation by $(ax-1)(bx-1),$ you get $$abx -a + abx - b = (a+b)\left(abx^2 - (a+b)x + 1\right)$$ you simplify this to $$ab(a+b)x^2 -x\left(a^2+ 4ab + b^2\right) + 2(a+b) = 0$$ the discriminant is

$$\begin{align} disc &= \left(a^2+ 4ab + b^2\right)^2 - 8ab(a+b)^2 \\ &= a^4 + 16a^2b^2 + b^4 + 8a^3b + 2a^2b^2 + 8ab^3 - \left( 8ab(a^2+2ab+b^2) \right)\\&= (a^2 + b^2)^2\end{align}$$

now, $$x = \frac{a^2 + 4ab + b^2 \pm(a^2 + b^2)}{2ab(a+b)}=\frac{a+b}{ab}, \frac 2{a+b} $$