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

1 Answer
Aug 26, 2015

Answer:

#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.