Question

Express with radical signs instead of fractional exponents. Rationalize the denominator.

Answer 1

$x^{\frac{1}{2}} = \sqrt[2]{x}$ and $3^{-\frac{1}{2}} = \frac{1}{3^{\frac{1}{2}} } = \frac{1}{\sqrt[2]{3}}$ so

$3^{-\frac{1}{2}} x^{\frac{1}{2}} = \frac{1}{\sqrt{3}} \cdot \sqrt{x} = \frac{\sqrt{x}}{\sqrt{3}}$

and you rationalize denominator by multiplying numerator and denominatr by $\sqrt{3}$ so that gives--

$\frac{\sqrt{x}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3x}}{3}$

Your answer is $\frac{\sqrt{3x}}{3}$