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AysviL [449]
3 years ago
11

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

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

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}

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