Fractional Exponents and Radicals: Practice

Simplify each radical, then combine the like terms. Here $75=25\cdot 3$ and $12=4\cdot 3$.

$\sqrt{75}+\sqrt{12}=5\sqrt{3}+2\sqrt{3}=7\sqrt{3}$

Roots do not split across a sum, so it is not $\sqrt{87}$.

Pull out the largest perfect-cube factor of $16$, which is $8$, since $16=8\cdot 2$.

$\sqrt[3]{16}=\sqrt[3]{8}\,\sqrt[3]{2}=2\sqrt[3]{2}$

Square the base first, which makes it positive, then take the non-negative root.

$\sqrt{(-6)^2}=\sqrt{36}=6$

The square and the root do not simply cancel to $-6$, because $\sqrt{a^2}$ returns the non-negative value.

Write each root as a one-half power and add the exponents.

$\sqrt{a}\cdot\sqrt{a}=a^{1/2}\cdot a^{1/2}=a^{1}=a$

Take the reciprocal, then read $32^{3/5}$ as a fifth root followed by a cube.

$32^{-3/5}=\frac{1}{\left(\sqrt[5]{32}\right)^3}=\frac{1}{2^3}=\frac{1}{8}$

A quotient of roots is the root of the quotient, so divide under a single radical.

$\frac{\sqrt{50}}{\sqrt{2}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5$

Write the cube root as a one-third power and multiply the exponents.

$\left(\sqrt[3]{5}\right)^6=\left(5^{1/3}\right)^6=5^{6/3}=5^2=25$

The cube root is a one-third power, and the reciprocal negates the exponent.

$\frac{1}{\sqrt[3]{a}}=\frac{1}{a^{1/3}}=a^{-1/3}$

An even-index root needs a radicand that is zero or positive, because no real number raised to an even power is negative.

$\sqrt[4]{-16}\ \text{is not a real number}$

The other three radicands are zero or positive, so their fourth roots are real.

Multiply the coefficients and multiply the radicals separately, using $\sqrt{3}\cdot\sqrt{3}=3$.

$5\sqrt{3}\cdot 2\sqrt{3}=(5\cdot 2)(\sqrt{3}\cdot\sqrt{3})=10\cdot 3=30$

The denominator $4$ is the index and the numerator $3$ is the inside power.

$x^{3/4}=\sqrt[4]{x^3}$

Swapping them to $\sqrt[3]{x^4}$ gives $x^{4/3}$, a different power.

Write the square root as a one-half power and multiply the exponents.

$\left(\sqrt{7}\right)^4=\left(7^{1/2}\right)^4=7^{4/2}=7^2=49$