Question

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

Answer 1

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`

Answer 2

`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`