Question

rationalize the denominator of (sqrt4{\frac{16}{25x^2}}). assume that all variables represent positive numbers. (sqrt4{\frac{16}{25x^2}} = square)

Explanation:

Step1: Rewrite root as fraction of roots

$\sqrt[4]{\frac{16}{25x^2}} = \frac{\sqrt[4]{16}}{\sqrt[4]{25x^2}}$

Step2: Simplify numerator

$\sqrt[4]{16} = \sqrt[4]{2^4} = 2$

Step3: Rewrite denominator with exponents

$\sqrt[4]{25x^2} = \sqrt[4]{5^2 x^2} = (5^2 x^2)^{\frac{1}{4}} = 5^{\frac{1}{2}}x^{\frac{1}{2}} = \sqrt{5x}$

Step4: Rationalize the denominator

Multiply numerator and denominator by $\sqrt[4]{25x^2}$:
$\frac{2}{\sqrt{5x}} \times \frac{\sqrt[4]{25x^2}}{\sqrt[4]{25x^2}} = \frac{2\sqrt[4]{25x^2}}{5x}$
Simplify $\sqrt[4]{25x^2} = \sqrt{5x}$, so:
$\frac{2\sqrt{5x}}{5x}$

Answer:

$\frac{2\sqrt{5x}}{5x}$