# Question #32f0e

Aug 19, 2017

$x = 0$

#### Explanation:

$\setminus \sqrt{60} = \setminus \sqrt{{2}^{2} \cdot 15} = 2 \setminus \sqrt{15}$

$\setminus \sqrt{15} = \setminus \sqrt{15}$

$\setminus \sqrt{135} = \setminus \sqrt{{3}^{2} \cdot 15} = 3 \setminus \sqrt{15}$

Now, we know that if two radicals have the same value underneath the square root, we can combine them, like so:

$a \setminus \sqrt{b} - c \sqrt{b} = \left(a - c\right) \sqrt{b}$

So,

$2 \setminus \sqrt{15} + \sqrt{15} - 3 \sqrt{15} = 3 \sqrt{15} - 3 \sqrt{15} = 0$

Thus,

$0 = \setminus \sqrt{x}$ so $x = 0$.