Question

Take a factor out of the square root:

Answer 1

Answer:

Question A)

$=\sqrt{6}x$

Question B)

$=3a\sqrt{a}$

Question C)

$=5\sqrt{2}b^2$

Step-by-step explanation:

A)

We are given:

$\sqrt{6x^2}\, \text{ where } x\geq 0$

We can rewrite the expression:

$=\sqrt{6}\cdot \sqrt{x^2}$

The square root and square will cancel each other out. Thus:

$=\sqrt{6}x$

B)

We are given:

$\sqrt{9a^3}$

Rewrite:

$=\sqrt{9}\cdot \sqrt{a^3}$

Note that the square root of 9 is simply 3. We can also factor the second part:

$=3\cdot \sqrt{a^2\cdot a}$

Rewriting:

$=3\cdot\sqrt{a^2}\cdot\sqrt{a}$

Simplify:

$=3a\sqrt{a}$

C)

We are given:

$\sqrt{50b^4}$

Rewrite. Note that 50 = 25(2):

$=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}$

Simplify. We can rewrite the factor as:

$=5\cdot \sqrt{2}\cdot \sqrt{(b^2)^2}$

The square and square root will cancel out. Thus:

$=5\sqrt{2}b^2$