Question

Which choice is equivalent to the product below when x>0?

Answer 1

Answer:

D

Step-by-step explanation:

Using the rule of radicals

$\sqrt{\frac{a}{b} }$ = $\frac{\sqrt{a} }{\sqrt{b} }$

$\sqrt{a\\}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Given

$\sqrt{\frac{6}{x} }$ × $\sqrt{\frac{x^2}{24} }$

= $\frac{\sqrt{6} }{\sqrt{x} }$ × $\frac{x^2}{24}$

= $\frac{\sqrt{6} }{\sqrt{x} }$ × $\frac{x}{2 \sqrt{6\\} }$

Cancel $\sqrt{6}$ on numerator/ denominator

= $\frac{1}{\sqrt{x} }$ × $\frac{x}{2\\}$

= $\frac{1}{\sqrt{x} }$ × $\frac{(\sqrt{x})^2 }{2}$

Cancel $\sqrt{x}$ on numerator/ denominator, leaving

= $\frac{\sqrt{x} }{2}$ → D