TITLE: Precalculus equation solving involving square roots
QUESTION [5 upvotes]: Solve the equation:
$$\sqrt x = \frac{3}{\sqrt x}+ \sqrt {x+3}$$
My approach was to multiply both sides with $\sqrt x$:
$$x = 3 + \sqrt {x+3} \sqrt x$$
$$(x - 3)^2 = (x+3)x$$
$$x^2 - 6x + 9 = 3x + x^2$$
$$9x-9 = 0$$
$$x = 1$$
...but this is clearly a false solution since 1 $\neq$ 5.
I sort of run out of steam here and have no more ideas. Is there a mistake in the math or does this just mean that there are no solutions (since the only solution that can be reached is a false one that was introduced by squaring)?

REPLY [6 votes]: Your algebra is correct and it does mean no solution exists. You could have in fact concluded no solution exists directly as follows:
(i) For all $x\geq 0$, we have $\sqrt{x+3} > \sqrt{x}$ and $\dfrac3{\sqrt{x}} > 0$.
(ii) Hence, $\sqrt{x+3} + \dfrac3{\sqrt{x}} > \sqrt{x}$ for all $x \geq 0$.
(iii) Hence, equality can never occur.