Algebra II - Understanding Radicals

Which of the following is equivalent to $2x^{\frac{3}{8}}$ ?

$2\sqrt[8]{x^3}$

$\sqrt[8]{2x^3}$

$2\sqrt[3]{x^8}$

$\sqrt[3]{2x^8}$

$\textup{The answer is not given.}$

Explanation

The denominator of the powered term represents the root of the radical. The numerator of the fraction is the power that the quantity of the radical is raised by.

Note that the integer in front of the x-term will not be included inside the radical.

The answer is: $2\sqrt[8]{x^3}$

Rewrite the following radical as an exponent:

$\sqrt[3]{y^{3}z^{7}}$

$yz^{\frac{7}{3}}$

$yz^{7}$

$yz^{\frac{1}{3}}$

Explanation

In order to rewrite a radical as an exponent, the number in the radical that indicates the root, gets written as a fractional exponent. Distribute the exponent to both terms by multiplying it by the exponents of each term as shown below:

$\sqrt[3]{y^{3}z^{7}}=(y^{3}z^{7})^{\frac{1}{3}}=y^{\frac{3}{3}}z^{\frac{7}{3}}$

From this point simplify the exponents accordingly:

$y^{\frac{3}{3}}z^{\frac{7}{3}}=yz^{\frac{7}{3}}$

Rewrite the radical as an exponent:

$\sqrt[5]{x^{4}}$

$x^{\frac{4}{5}}$

$x^{\frac{1}{4}}$

$x^{\frac{5}{4}}$

$x^{\frac{1}{5}}$

Explanation

In order to rewrite a radical as an exponent, the number in the radical that indicates the root, gets written as a fractional exponent as shown below:

$\sqrt[5]{x^{4}}=x^{\frac{4}{5}}$

In this case, we are done because there are no further simplification steps.

Evaluate the radical: $\sqrt{8}\times \sqrt{40} \times \sqrt{24}$

$16\sqrt{30}$

$12\sqrt{30}$

$14\sqrt{30}$

$18\sqrt{30}$

$11\sqrt{21}$

Explanation

In order to get the most simplified answer, do not multiply all the numbers together and combine as one radical.

Rewrite each radicals in their most simplified form.

$\sqrt8 = \sqrt4 \cdot \sqrt2 = 2\sqrt2$

$\sqrt{40} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10}$

$\sqrt{24}= \sqrt{4}\cdot \sqrt 6 = 2\sqrt6$

Multiply the terms.

$2\sqrt{2} \cdot 2\sqrt{10} \cdot2\sqrt{6} = (2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

Notice that multiplying a square root of a number by itself will leave only the integer and will eliminate the radical.

$\sqrt2 \cdot \sqrt2 =2$

The terms become:

$=(2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

$= 2\cdot2\cdot2\cdot2 (\sqrt2 \cdot \sqrt5\cdot \sqrt3)$

$=16\sqrt{30}$

The answer is: $16\sqrt{30}$

Evaluate: $(\frac{2}{3})^{\frac{1}{2}}$

$\frac{\sqrt6}{3}$

$2\sqrt6$

$\sqrt3$

$\frac{\sqrt2}{6}$

$\frac{\sqrt2}{3}$

Explanation

The exponent can be distributed through the numerator and denominator.

$(\frac{2}{3})^{\frac{1}{2}}=\frac{2^{\frac{1}{2}}}{3^{\frac{1}{2}}}$

Rewrite both the top and bottom with radicals. The power of one-half is the same as taking the square root of the number.

$\frac{\sqrt2}{\sqrt3}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\frac{\sqrt2}{\sqrt3}\cdot \frac{\sqrt3}{\sqrt3} =\frac{\sqrt6}{3}$

The answer is: $\frac{\sqrt6}{3}$

Rewrite the following radical as an exponent:

$\sqrt[3]{y^{3}z^{7}}$

$yz^{\frac{7}{3}}$

$yz^{7}$

$yz^{\frac{1}{3}}$

Explanation

$\sqrt[3]{y^{3}z^{7}}=(y^{3}z^{7})^{\frac{1}{3}}=y^{\frac{3}{3}}z^{\frac{7}{3}}$

From this point simplify the exponents accordingly:

$y^{\frac{3}{3}}z^{\frac{7}{3}}=yz^{\frac{7}{3}}$

Which of the following is equivalent to $2x^{\frac{3}{8}}$ ?

$2\sqrt[8]{x^3}$

$\sqrt[8]{2x^3}$

$2\sqrt[3]{x^8}$

$\sqrt[3]{2x^8}$

$\textup{The answer is not given.}$

Explanation

The denominator of the powered term represents the root of the radical. The numerator of the fraction is the power that the quantity of the radical is raised by.

Note that the integer in front of the x-term will not be included inside the radical.

The answer is: $2\sqrt[8]{x^3}$

Rewrite the radical as an exponent:

$\sqrt[5]{x^{4}}$

$x^{\frac{4}{5}}$

$x^{\frac{1}{4}}$

$x^{\frac{5}{4}}$

$x^{\frac{1}{5}}$

Explanation

In order to rewrite a radical as an exponent, the number in the radical that indicates the root, gets written as a fractional exponent as shown below:

$\sqrt[5]{x^{4}}=x^{\frac{4}{5}}$

In this case, we are done because there are no further simplification steps.

Evaluate the radical: $\sqrt{8}\times \sqrt{40} \times \sqrt{24}$

$16\sqrt{30}$

$12\sqrt{30}$

$14\sqrt{30}$

$18\sqrt{30}$

$11\sqrt{21}$

Explanation

In order to get the most simplified answer, do not multiply all the numbers together and combine as one radical.

Rewrite each radicals in their most simplified form.

$\sqrt8 = \sqrt4 \cdot \sqrt2 = 2\sqrt2$

$\sqrt{40} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10}$

$\sqrt{24}= \sqrt{4}\cdot \sqrt 6 = 2\sqrt6$

Multiply the terms.

$2\sqrt{2} \cdot 2\sqrt{10} \cdot2\sqrt{6} = (2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

Notice that multiplying a square root of a number by itself will leave only the integer and will eliminate the radical.

$\sqrt2 \cdot \sqrt2 =2$

The terms become:

$=(2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

$= 2\cdot2\cdot2\cdot2 (\sqrt2 \cdot \sqrt5\cdot \sqrt3)$

$=16\sqrt{30}$

The answer is: $16\sqrt{30}$

Evaluate: $(\frac{2}{3})^{\frac{1}{2}}$

$\frac{\sqrt6}{3}$

$2\sqrt6$

$\sqrt3$

$\frac{\sqrt2}{6}$

$\frac{\sqrt2}{3}$

Explanation

The exponent can be distributed through the numerator and denominator.

$(\frac{2}{3})^{\frac{1}{2}}=\frac{2^{\frac{1}{2}}}{3^{\frac{1}{2}}}$

Rewrite both the top and bottom with radicals. The power of one-half is the same as taking the square root of the number.

$\frac{\sqrt2}{\sqrt3}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\frac{\sqrt2}{\sqrt3}\cdot \frac{\sqrt3}{\sqrt3} =\frac{\sqrt6}{3}$

The answer is: $\frac{\sqrt6}{3}$

Understanding Radicals

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