# How do you solve and check for extraneous solutions in (-2x+6)^(1/5)=(-8+10x)^(1/5)?

Aug 26, 2015

$x = \frac{7}{6}$

#### Explanation:

Actually, your equation will not have extraneous solutions because you can have negative numbers under the radicals.

Rewrite your equation to radical form to get a better understanding of what's going on

$\sqrt[5]{- 2 x + 6} = \sqrt[5]{- 8 + 10 x}$

Since you're dealing with odd-indexed radicals, you don't have to worry about the sign of the two expressions that are under the radicals.

Remember that when you multiply a negative number by itself an odd number of times, you still get a negative number.

So, raise both sides of the equation to the fifth power

${\left(\sqrt[5]{- 2 x + 6}\right)}^{5} = {\left(\sqrt[5]{- 8 + 10 x}\right)}^{5}$

$- 2 x + 6 = - 8 + 10 x$

Now simply rearrange and solve for $x$

$12 x = 14 \implies x = \frac{14}{12} = \textcolor{g r e e n}{\frac{7}{6}}$

You can check this solution using a calculator

$\sqrt[5]{- 2 \cdot \frac{7}{6} + 6} = 1.296744116 \ldots$

and

$\sqrt[5]{- 8 + 10 \cdot \frac{7}{6}} = 1.296744116 \ldots$