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

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Answer 1 · 2015-08-26T22:45:50

#x = 7/6#

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

#root(5)(-2x+6) = root(5)(-8+10x)#

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

#(root(5)(-2x+6))^5 = (root(5)(-8 + 10x))^5#

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

Now simply rearrange and solve for #x#

#12x = 14 implies x = 14/12 = color(green)(7/6)#

You can check this solution using a calculator

#root(5)(-2 * 7/6 + 6) = 1.296744116...#

and

#root(5)(-8 + 10 * 7/6) = 1.296744116...#