# How do you simplify sqrt(1+x) - sqrt(1-x)?

Sep 29, 2015

$\sqrt{1 + x} - \sqrt{1 - x}$ does not really simplify,

but you can re-express it in various ways...

#### Explanation:

First note that for both square roots to have Real values, we must have $x \in \left[- 1 , 1\right]$

Let's see what happens when you square $\sqrt{1 + x} - \sqrt{1 - x}$ ...

${\left(\sqrt{1 + x} - \sqrt{1 - x}\right)}^{2}$

$= {\left(\sqrt{1 + x}\right)}^{2} - 2 \left(\sqrt{1 + x}\right) \left(\sqrt{1 - x}\right) + {\left(\sqrt{1 - x}\right)}^{2}$

$= \left(1 + x\right) - 2 \sqrt{1 - {x}^{2}} + \left(1 - x\right)$

[[ using $\sqrt{a} \sqrt{b} = \sqrt{a b}$ ]]

$= 2 - 2 \sqrt{1 - {x}^{2}}$

So $\sqrt{1 + x} - \sqrt{1 - x} = \pm \sqrt{2 - 2 \sqrt{1 - {x}^{2}}}$

What is the correct sign to choose?

If $x \ge 0$ then $1 + x \ge 1 - x$, so $\sqrt{1 + x} - \sqrt{1 - x} \ge 0$

If $x < 0$ then $1 + x < 1 - x$, so $\sqrt{1 + x} - \sqrt{1 - x} < 0$

So we have:

$\sqrt{1 + x} - \sqrt{1 - x} = \left\{\begin{matrix}\sqrt{2 - 2 \sqrt{1 - {x}^{2}}} & \text{if " x >= 0 \\ -sqrt(2 - 2sqrt(1-x^2)) & "if } x < 0\end{matrix}\right.$

If you like, you can separate out the common factor $2$ to get:

$\sqrt{1 + x} - \sqrt{1 - x} = \left\{\begin{matrix}\sqrt{2} \sqrt{1 - \sqrt{1 - {x}^{2}}} & \text{if " x >= 0 \\ -sqrt(2)sqrt(1 - sqrt(1-x^2)) & "if } x < 0\end{matrix}\right.$

graph{sqrt(1+x)-sqrt(1-x) [-5, 5, -2.5, 2.5]}