Checkpoint: Equations with Fractional Exponents Solution

Solve. Check your work.
$\frac{5}{3} {(4 - x)}^{\frac{1}{5}} + 5 = 10$

$\begin{array}{rcl} \frac{5}{3} {(4 - x)}^{\frac{1}{5}} & = & 5 \\ {(4 - x)}^{\frac{1}{5}} & = & 3 \\ {({(4 - x)}^{\frac{1}{5}})}^{5} & = & {(3)}^{5} \\ 4 - x & = & 243 \\ - x & = & 239 \\ x & = & - 239 \end{array}$

Check
$\begin{array}{rcl} \frac{5}{3} {(4 - (- 239))}^{\frac{1}{5}} + 5 & = & 10 \\ \frac{5}{3} {(243)}^{\frac{1}{5}} + 5 & = & 10 \\ \frac{5}{3} (3) + 5 & = & 10 ✓ \end{array}$

Note

Q: Will all negative values be extraneous for radical equations?

A: No, the value can be negative as long as both sides of the equation are equal.