Make x the subject given the formula for y

Question

I am given the following formula:

$$y=\frac{x}{a}+\sqrt{\frac{x}{b}}.$$

I want to make x the subject.

I rearranged the equation and got to:

$$y^{4}=x(\frac{y^{2}}{2}+2y^{2})-x^{2}.$$

and I don't know where to go from here. May be this is the wrong rearrangement.

The answer according to wolfram alpha is:

$$x=\pm \frac{a^{\frac{3}{2}}\sqrt{a+4by}+a^{2}+2aby}{2b}.$$

How can I get there?

Answer 1

See $y$ as a quadratic function of $\sqrt {x}$. Use the quadratic formula.

Answer 2

$\textbf{a different approach}$

let $x = t^2$ thus your equation becomes $$ y = \dfrac{t^2}{a} + \dfrac{t}{\sqrt{b}} $$ solve for $t$ then compute $x$.

Answer 3

Hint: Set $u=y^2$ and solve the quadratic equation for $u$.

Then your equation is rewritten as $$ u^2=\frac{5x}{2}u-x^2, $$

which by completing the square is written as

$$ \Bigg(u-\frac{5}{4}x\Bigg)^2 - \frac{9}{16}x^2=0, $$

and thus you have

$$ u-\frac{5}{4}x = \pm\frac{3}{4}x, $$

or

$$ u = \pm\frac{3}{4}x+\frac{5}{4}x = \frac{5\pm3}{4}x. $$ Finally $y=\sqrt{u}=\sqrt{\frac{5\pm3}{4}x}$.