# 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.

## 5 Answers

\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}

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.

Expand and solve:

$x = \frac{a^4-2 a^2 b-2 a^2 c+b^2-2 b c+c^2}{4 a^2} = \frac{a^4-2 a^2 (b+c)+(b-c)^2}{4 a^2}$

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*}

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 $$b or $$b>c$$ but it doesn't really matter.