How to isolate $a$ in this equation?

Question

I have the following equation that I would like to isolate the variable $a$ from, but I am a bit stuck. The variable $k$ is known; however, $x$ and $a$ are unknowns.

$$ x^2 + \frac{a}{(1-a)} x -\frac{1}{2(1-a)k}=0 $$

Upon solving for $x$ we find

$$ x = \frac 12 \bigg\{-\frac{a}{(1-a)}\pm \bigg[\frac{a^2}{(1-a)^2}+\frac{2}{(1-a)k}\bigg]^{1/2}\bigg\} $$

At this point I think that we can substitute this solution into the first equation to obtain

$$ \frac 14 \bigg\{-\frac{a}{(1-a)}\pm \bigg[\frac{a^2}{(1-a)^2}+\frac{2}{(1-a)k}\bigg]^{1/2}\bigg\}^2 + \frac{a}{2(1-a)} \bigg\{-\frac{a}{(1-a)}\pm \bigg[\frac{a^2}{(1-a)^2}+\frac{2}{(1-a)k}\bigg]^{1/2}\bigg\} -\frac{1}{2(1-a)k}=0 $$

Can this expression be inverted to give $a=?$, and if so, how can I achieve this?

Answer 1

The equation is $x^2-x+\frac{x}{1-a}-\frac1{2k(1-a)}=0$, therefore $$\frac1{1-a}\left(x-\frac1{2k}\right) =x-x^2\\ a=1-\frac{2kx-1}{2kx(1-x)}$$ for $k\ne0$, $x\notin\{ 0,1,\frac1{2k}\}$. If $k=\frac12$ and $x=1$, then every $a\ne 1$ satisfies the equation. In all other cases either the equation does not exist or no $a$ satisfies it.

Answer 2

Multiply the first equation by $1-a$ to get $$(1-a)x^2+ax-\frac{1}{2k}=0.$$ This is linear in $a$; collecting like terms yields $$(x-x^2)a+x^2-\frac{1}{2k}=0,$$ and hence if $x\notin\{0,1\}$ then $$a=\frac{\tfrac{1}{2k}-x^2}{x-x^2}=\frac{1-2kx^2}{2kx-2kx^2}.$$

Answer 3

Always if you substitute the solution in the equation, you just get a tautology.

You must solve it according to $a$ instead of $x$, as @Gae. S. solved it for you.