Solve frac{sqrt[4]{36}sqrt[4]{9}}{sqrt[4]{4}}

$$\frac{ \sqrt[ 4 ]{ 36 } \sqrt[ 4 ]{ 9 } }{ \sqrt[ 4 ]{ 4 } }$$

Evaluate

$$3$$

Solution Steps

Rewrite \sqrt[4]{36} as \sqrt[4]{6^{2}}. Convert from radical to exponential form and cancel out 2 in the exponent. Convert back to radical form.

$$\sqrt[4]{36}=\sqrt[4]{6^{2}}=6^{\frac{2}{4}}=6^{\frac{1}{2}}=\sqrt{6}$$

Insert the obtained value back in the expression.

$$\frac{\sqrt{6}\sqrt[4]{9}}{\sqrt[4]{4}}$$

Rewrite \sqrt[4]{9} as \sqrt[4]{3^{2}}. Convert from radical to exponential form and cancel out 2 in the exponent. Convert back to radical form.

$$\sqrt[4]{9}=\sqrt[4]{3^{2}}=3^{\frac{2}{4}}=3^{\frac{1}{2}}=\sqrt{3}$$

Insert the obtained value back in the expression.

$$\frac{\sqrt{6}\sqrt{3}}{\sqrt[4]{4}}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{\sqrt{3}\sqrt{2}\sqrt{3}}{\sqrt[4]{4}}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{3\sqrt{2}}{\sqrt[4]{4}}$$

Rewrite \sqrt[4]{4} as \sqrt[4]{2^{2}}. Convert from radical to exponential form and cancel out 2 in the exponent. Convert back to radical form.

$$\sqrt[4]{4}=\sqrt[4]{2^{2}}=2^{\frac{2}{4}}=2^{\frac{1}{2}}=\sqrt{2}$$

Insert the obtained value back in the expression.

$$\frac{3\sqrt{2}}{\sqrt{2}}$$

Rationalize the denominator of \frac{3\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{3\sqrt{2}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{2}\sqrt{2}}{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{3\times 2}{2}$$

Multiply 3 and 2 to get 6.

$$\frac{6}{2}$$

Divide 6 by 2 to get 3.

$$3$$

Factor

$$3$$