# Can you write (a+b)^0.5 - (a-b)^0.5 as the square root of a difference?

Mar 4, 2018

${\left(a + b\right)}^{0.5} - {\left(a - b\right)}^{0.5} = \sqrt{2 a - 2 \sqrt{{a}^{2} - {b}^{2}}}$

#### Explanation:

If ${\left(a + b\right)}^{0.5} - {\left(a - b\right)}^{0.5}$ a.k.a. $\sqrt{a + b} - \sqrt{a - b}$ can be written as the square root of a difference, then squaring it should give us a difference...

${\left(\sqrt{a + b} - \sqrt{a - b}\right)}^{2} = {\left(\sqrt{a + b}\right)}^{2} - 2 \sqrt{a + b} \sqrt{a - b} + {\left(\sqrt{a - b}\right)}^{2}$

$\textcolor{w h i t e}{{\left(\sqrt{a + b} - \sqrt{a - b}\right)}^{2}} = \left(a + b\right) - 2 \sqrt{a + b} \sqrt{a - b} + \left(a - b\right)$

$\textcolor{w h i t e}{{\left(\sqrt{a + b} - \sqrt{a - b}\right)}^{2}} = 2 a - 2 \sqrt{\left(a + b\right) \left(a - b\right)}$

$\textcolor{w h i t e}{{\left(\sqrt{a + b} - \sqrt{a - b}\right)}^{2}} = 2 a - 2 \sqrt{{a}^{2} - {b}^{2}}$

So yes, ${\left(a + b\right)}^{0.5} - {\left(a - b\right)}^{0.5}$ can be written as the square root of the difference of $2 a$ and $2 \sqrt{{a}^{2} - {b}^{2}}$, namely:

${\left(a + b\right)}^{0.5} - {\left(a - b\right)}^{0.5} = \sqrt{2 a - 2 \sqrt{{a}^{2} - {b}^{2}}}$

Mar 4, 2018

$\sqrt{2 a - 2 \sqrt{{a}^{2} - {b}^{2}}}$

#### Explanation:

$\sqrt{x - y} = \sqrt{a + b} - \sqrt{a - b}$

squaring both sides

$x - y = 2 a - 2 \sqrt{{a}^{2} - {b}^{2}}$ and then

$\sqrt{x - y} = \sqrt{a + b} - \sqrt{a - b} = \sqrt{2 a - 2 \sqrt{{a}^{2} - {b}^{2}}}$