# Precalculus equation solving involving square roots

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)?

• I don't see any mistakes. And you are right to assume there are no solutions. – randomgirl Apr 22 '15 at 21:23
• This does indeed show that the equation has no solutions, since you have shown that if $x$ is a solution, then $x=1$. But since $x=1$ isn't a solution, there are no solutions. – Uncountable Apr 22 '15 at 21:26

(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 $x \geq 0$.