Fractional Exponents and Radicals: Additional Practice (Set 1)

Take the cube root (denominator $3$), then square (numerator $2$).

$125^{2/3}=\left(\sqrt[3]{125}\right)^2=5^2=25$

Pull out the largest perfect-square factor of $98$, which is $49$, since $98=49\cdot 2$.

$\sqrt{98}=\sqrt{49}\,\sqrt{2}=7\sqrt{2}$

Simplify $\sqrt{32}$ first, then combine like radicals. Since $32=16\cdot 2$, $\sqrt{32}=4\sqrt{2}$.

$\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}$

Roots do not split across a sum, so it is not $\sqrt{34}$.

Write each root as a power, then add the exponents.

$a^{2/3}\cdot a^{1/3}=a^{\frac{2}{3}+\frac{1}{3}}=a^{1}=a$

Pull out the largest perfect-cube factor of $250$, which is $125$, since $250=125\cdot 2$.

$\sqrt[3]{250}=\sqrt[3]{125}\,\sqrt[3]{2}=5\sqrt[3]{2}$

A cube root of a fraction takes the cube root of the top and bottom separately.

$\left(\tfrac{8}{27}\right)^{1/3}=\frac{\sqrt[3]{8}}{\sqrt[3]{27}}=\frac{2}{3}$

A square root has an even index, so its radicand must be zero or positive.

$\sqrt{-9}\ \text{is not a real number}$

The other three radicands are zero or positive.

Take the reciprocal, then read $9^{3/2}$ as a square root followed by a cube.

$9^{-3/2}=\frac{1}{\left(\sqrt{9}\right)^3}=\frac{1}{3^3}=\frac{1}{27}$

Simplify each radical, then combine. Here $28=4\cdot 7$ and $63=9\cdot 7$.

$\sqrt{28}+\sqrt{63}=2\sqrt{7}+3\sqrt{7}=5\sqrt{7}$

A one-third power is a cube root, and a cube root has an odd index, so a negative base is allowed.

$(-8)^{1/3}=\sqrt[3]{-8}=-2$

Check: $(-2)^3=-8$.

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

$16^{3/4}=\left(\sqrt[4]{16}\right)^3=2^3=8$

The exponent is not a multiplier, so it is not $16\cdot\tfrac{3}{4}$.

A negative exponent flips the fraction, turning the base into $8$; then evaluate $8^{2/3}$.

$\left(\tfrac{1}{8}\right)^{-2/3}=8^{2/3}=\left(\sqrt[3]{8}\right)^2=2^2=4$