Solve for x in terms of y, given $y=\frac{x^2}{x+1}$.

Question

I attempted to solve for x, but don't know how I should. Thanks for your help.

Answer 1

Multiply both sides by $1+x$ to get

$$(1+x) y = x^2$$

which is

$$y + xy = x^2 \implies x^2 - yx - y = 0$$

Answer 2

Writing out $$x^2-yx-y=0$$ you then have $$x_{1,2}=\frac{y\pm \sqrt{y^2+4y}}{2}$$ Of course you need to have the condition that $y^2 \geq -4y$ for this to have solutions for $x\in\mathbb{R}$. Hence if $y\geq 0$ you have that you have solutions for $y\geq-4$, which of course is always true, as you already had $y\geq0$. And finally if $y<0$, then you have $y\leq-4$.

Second Method: Write $$x^2-yx = y$$ Add $\frac{y^2}{4}$ to both sides $$x^2-yx +\frac{y^2}{4} = y+\frac{y^2}{4}$$ Rewrite the square as follows $$\left(x-\frac{y}{2}\right)^2=y+\frac{y^2}{4}$$ Take the square root both sides and get $$x-\frac{y}{2} = \pm\sqrt{\frac{y^2}{4}+y}$$ Hence you have $$x = \frac{y}{2}\pm\sqrt{\frac{y^2}{4}+y}$$

Which is indeed equivalent to the one above, just in a slightly more elegant way.