How do you solve # (x-2)^(3/4)=8#?

2 Answers
Mar 25, 2016

#(x-2)^(3/4)=8#

There is an easy solution which is:

#((x-2)^(3/4))^(4/3)=8^(4/3)#

#(x-2)^1=(2^3)^(4/3)#

#x-2=2^4=16#

#x=18#,

But we can get other solutions:

#(x-2)^(3/4)=8#

#((x-2)^(3/4))^4=8^4# This step is dangerous since we can be adding false solutions.

#(x-2)^3=8^4=4096#

#x^3-6x^2+12x-8=4096#

#x^3-6x^2+12x-4104=0#

We already know that 18 is one of the zeros:

So we divide the third degree polynom by (x-18):

#(x^3-6x^2+12x-4104)/(x-18)=x^2+12x+228#

#x^2+12x+228# can be solved by the quadratic formula:

#x=(-12+-sqrt(12^2-4*228))/2#

#x=(-12+-sqrt(-768))/2#

#x=(-12+-16sqrt(-3))/2#

#x=-6+-8isqrt(3)#

These are false solutions, since #(x-2)^(3/4)=+-8i#. These false solutions were added in the step where it is written :"This step is dangerous since we can be adding false solutions."

So the only solution is x=18.

Answer:

#x=18#

Assuming no complex solution is required

Explanation:

Given:#" "(x-2)^(color(magenta)(3/4))=8#

Remember that #(x-2)^(3/4)=root(3)((x-2)^4)#

Take the cube root of both sides giving

#" "(x-2)^(color(magenta)(1/4))= 2#

Raise both sides to the power of 4 giving

#x-2=2^4 = 16#

#x=18#