Question #32f0e

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Answer 1 · 2017-08-19T01:32:17

$x = 0$ $x=0$

Let's simplify the radicals.

$\sqrt{60} = \sqrt{2^{2} \cdot 15} = 2 \sqrt{15}$ $\sqrt(60)=\sqrt(2^2*15)=2\sqrt(15)$

$\sqrt{15} = \sqrt{15}$ $\sqrt(15)=\sqrt(15)$

$\sqrt{135} = \sqrt{3^{2} \cdot 15} = 3 \sqrt{15}$ $\sqrt(135)=\sqrt(3^2*15)=3\sqrt(15)$

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

$a \sqrt{b} - c \sqrt{b} = (a - c) \sqrt{b}$ $a\sqrt(b)-csqrt(b)=(a-c)sqrt(b)$

So,

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

Thus,

$0 = \sqrt{x}$ $0=\sqrt(x)$ so $x = 0$ $x=0$ .