How do you solve #root3(x^2)= 9#?

2 Answers
Mar 30, 2017

Answer:

Undo each of the things done to #x# one at a time.

Explanation:

First, to undo the cube root, we can cube both sides to get

#x^2=729#

Next, to undo the square, we can take the square root of both sides to get

#x=+-27#

Mar 30, 2017

Answer:

#x=+-27#

Explanation:

Given:

#root(3)(x^2)=9#

Note that #9=3^2#, so this can also be written:

#root(3)(x^2) = 3^2#

Note that both #t rarr t^3# and its inverse #t rarr root(3)(t)# are one to one as functions of real numbers. So if we cube both sides of the equation then we neither lose any solution nor introduce any extraneous solutions:

Cubing both sides of the equation, we get:

#x^2 = (3^2)^3 = 3^(2*3) = 3^(3*2) = (3^3)^2 = 27^2#

Subtract #27^2# from both ends to get:

#x^2-27^2 = 0#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Using this with #a=x# and #b=27# we find:

#0 = x^2-27^2 = (x-27)(x+27)#

So:

#x = +-27#