Question

How do you solve and check for extraneous solutions in `2(x + 8) ^ (4/5) - 12 = 150`?

Answer 1

`x = 235 " "` or `" "x = -251`

Explanation:

Start by rewriting your equation using the radical form for the term that has the fractional exponent

`2root(5)( (x+8)^4) - 12 = 150`

Next, isolate the radical term on the left-hand side of the equation by adding `12` to both sides and dividing all the terms by `2`

`2root(5)( (x+8)^4) - color(red)(cancel(color(black)(12))) + color(red)(cancel(color(black)(12))) = 150 + 12`

`(color(red)(cancel(color(black)(2)))root(5)( (x+8)^4))/color(red)(cancel(color(black)(2))) = 162/2`

`root(5)( (x+8)^4) = 81`

This can be rewritten as

`(root(5)(x+8))^4 = 81`

Take the fourth root from both sides of the equation - do not forget that you have positive and negative roots for the fourth root of `81`!

`root(4)((root(5)(x+8))^4) = root(4)(81)`

`root(5)(x+8) = +-3`

Raise both sides of the equation to the fifth power

`(root(5)(x+8))^5 = (+-3)^5`

This equation will now produce two solutions

`x+8 = 3^5`

`x = 243 - 8 = color(green)(235)`

and

`x + 8 = (-3)^5`

`x = -243 - 8 = color(green)(-251)`

Your original equation will thus have two valid solutions and no extraneous solutions.