# Question #4a68b

Oct 19, 2016

$x = \frac{15}{4}$

#### Explanation:

Note that as we have $4 - x$ under a radical, we must have $x \le 4$ to avoid taking the root of a negative number.

$4 + \sqrt{10 - x} = 6 + \sqrt{4 - x}$

$\implies \sqrt{10 - x} = 2 + \sqrt{4 - x}$

$\implies {\left(\sqrt{10 - x}\right)}^{2} = {\left(2 + \sqrt{4 - x}\right)}^{2}$

$\implies 10 - x = {2}^{2} + 2 \left(2\right) \sqrt{4 - x} + {\left(\sqrt{4 - x}\right)}^{2}$

$\implies 10 - x = 4 + 4 \sqrt{4 - x} + 4 - x$

$\implies 2 = 4 \sqrt{4 - x}$

$\implies \sqrt{4 - x} = \frac{1}{2}$

$\implies {\left(\sqrt{4 - x}\right)}^{2} = {\left(\frac{1}{2}\right)}^{2}$

$\implies 4 - x = \frac{1}{4}$

$\implies x = 4 - \frac{1}{4}$

$\therefore x = \frac{15}{4}$

Checking our result:

$4 + \sqrt{10 - \frac{15}{4}} = 4 + \sqrt{\frac{25}{4}}$

$= 4 + \frac{5}{2}$

$= \frac{13}{2}$

$= 6 + \frac{1}{2}$

$= 6 + \sqrt{\frac{1}{4}}$

$= 6 + \sqrt{4 - \frac{15}{4}}$

as desired