Solving for $x$ [not homework]

Question

How do I bring the remaining $x$ to the LHS?

$\pm x=\frac{(2-x)\sqrt{|q_2|} } {\sqrt{|q_1|}}$

to get

$x=\frac{2 \sqrt{|q_2|}}{\sqrt{|q_2|} \pm \sqrt{|q_1|}}$

I'm just not sure about the background operations to isolate the $x$.

Answer 1

Rewrite $$ \frac{(2-x)\sqrt{\lvert q_2\rvert}}{\sqrt{\lvert q_1\rvert}}=(2-x)\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}=2\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}-x\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}. $$ Then you can add the latter term to both sides of your equation to get $$ x\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}\pm x=2\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}, $$ or $$ x\left[\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}\pm1\right]=2\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}. $$ Dividing both sides by the second term on the left then yields $$ \begin{align*} x=\frac{2\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}}{\sqrt{\left\lvert\frac{q_2}{q_1}\right\rvert}\pm1}=\frac{2\sqrt{\lvert q_2\rvert}}{\sqrt{\lvert q_2\rvert}\pm\sqrt{\lvert q_1\rvert}}, \end{align*} $$ where the last step follows by multiplying both the numerator and denominator by $\sqrt{\lvert q_1\rvert}$.