solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging and then squaring doesn't help either.
The problem looks very simple to me but I have no idea on how to approach this.
 A: $$\begin{align}a=\sqrt{b+x}+\sqrt{c+x}&\Rightarrow \sqrt{b+x}=a-\sqrt{c+x}\\&\Rightarrow b+x=a^2-2a\sqrt{c+x}+c+x\\&\Rightarrow 2a\sqrt{c+x}=a^2+c-b\\&\Rightarrow 4a^2(c+x)=(a^2+c-b)^2\\&\Rightarrow x=\frac{(a^2+c-b)^2-4a^2c}{4a^2}\end{align}$$
A: Square once, and you have just one square root left $\sqrt{(b+x)(c+x)}$. Rearrange everything else to the the other side, square again, and there are no more roots, just a quadratic equation.
A: Expand and solve:
$x = \frac{a^4-2 a^2 b-2 a^2 c+b^2-2 b c+c^2}{4 a^2} = \frac{a^4-2 a^2 (b+c)+(b-c)^2}{4 a^2}$
A: HINT
If we agree that $\alpha = \sqrt{b + x}$ and $\beta = \sqrt{c + x}$ we obtain that $\alpha + \beta = a$ and $\alpha^{2} -\beta^{2} = b - c$. Hence it can be claimed $(\alpha + \beta)(\alpha - \beta) = a(\alpha - \beta) = b-c$ which implies the following system of equations
\begin{align*}
\begin{cases}
\alpha - \beta = \displaystyle\frac{b-c}{a}\\
\alpha + \beta = a
\end{cases} \Rightarrow 2\alpha  = a + \frac{b-c}{a} = \frac{a^{2} + b - c}{a}\Leftrightarrow \boxed{\alpha = \sqrt{b+x} = \frac{a^{2} + b - c}{2a}}
\end{align*}
A: To solve
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
you must square twice, since you have two square roots.
First, for $\sqrt{b + x}$ and $\sqrt{c + x}$ to both exist, it must be $x\ge-b$ and $x\ge-c$, so
$$
x\ge\max(-b,-c).
$$
Then:
$$
a - \sqrt{b + x} = \sqrt{c + x}.\tag{1}
$$
Squaring once we find
\begin{align}
& a^2 + b \color{red}{+ x} - 2a\sqrt{b + x} = c \color{red}{+ x}\\
& a^2 + b - c = 2a\sqrt{b + x}\\
\end{align}
and squaring the second time:
$$
(a^2 + b - c)^2 = 4a^2(b + x),
$$
so
$$
x=\frac{(a^2 + b - c)^2}{4a^2} - b
$$
or
$$
x=\left(\frac{a^2 + b - c}{2a}\right)^2 - b.
$$
If you rewrite $(1)$ as
$$
a - \sqrt{c + x} = \sqrt{b + x},
$$
you will find
$$
x=\frac{(a^2 + c - b)^2}{4a^2} - c
$$
or
$$
x=\left(\frac{a^2 + c - b}{2a}\right)^2 - c
$$
which is the same.
Here $a$ (not $x$) is a symmetric function of $b$ and $c$ so one may wonder if $b<c$ or $b>c$ but it doesn't really matter.
