$0$ as valid solution

Question

In the equation

$\frac {4}{x+ \sqrt{x^2+x}}-\frac {1}{x- \sqrt{x^2+x}} = \frac{3}{x}$

is $x=0$ considered a valid solution? The other one being $x=\frac{9}{16}$.

After all, if $x=0$ the whole equation ends as $0=0$.

Answer 1

No. If $x=0$, then what you get is $\frac40-\frac10=\frac30$, which is meaningless.

Answer 2

As noticed $0$ is not a solution, for that reason we can proceed as follows

$$\frac {4}{x+ \sqrt{x^2+x}}-\frac {1}{x- \sqrt{x^2+x}} = \frac{3}{x} \iff x\cdot\frac {4}{x+ \sqrt{x^2+x}}-x\cdot\frac {1}{x- \sqrt{x^2+x}} = x\cdot\frac{3}{x}$$

$$ \frac {4x}{x+ \sqrt{x^2+x}}-\frac {x}{x- \sqrt{x^2+x}} = 3$$

$$ \frac {4}{1+ \sqrt{1+1/x}}-\frac {1}{1- \sqrt{1+1/x}} = 3$$

and then solve by $t^2=1+1/x\ge 0$.