An equation involving radicals

While browsing the internet, the following brainteaser popped up on the screen: find $x$ such that

$$\sqrt{x+15} + \sqrt{x} = 15$$

A straight forward approach is to square both sides, move the single term with radicals to one side and square again, and reducing terms to obtain a linear equation in $x$ whose solution gives the answer.

Here is another (somewhat simpler) way to solve this problem. First note that the left hand side is a increasing function of $x$ and at $x=0$ is equal to $\sqrt{15} < 15$, so there is a single real solution to the equation above.
If we set $x = y^2$ and $x+15 = (y+z)^2$, we get $15 = 2yz + z^2 = z(2y+z)$. The left hand side of the original equation then becomes:

$y+z +y = 2y+z = 15$. Combine this with the above it follows that $z = 1$ and $y = 7$. Thus the answer is $x = 49$.

In general, this method shows that the equation

$$\sqrt{x+a} + \sqrt{x} = b$$ where $b \geq 0$ and $b^2 \geq a$ has as the only real solution:

$$x = \left(\frac{b^2-a}{2b}\right)^2$$

For the case $a = b \geq 0$, this reduces to

$$x = \left(\frac{a-1}{2}\right)^2$$

which is an integer if $a$ is odd.