Question

How do you solve `(x+5)^4 = 81`?

Answer 1

If we're only examining real numbers, the two solutions are `x=-2` and `x=-8`.

Explanation:

Take the `4`th root of both sides to undo the `4`th power on `(x+5)^4`.

`root4((x+5)^4)=root4 81`

Note that `81=3^4`. Thus, `root 4 81=+-3`. This is one of the trickier parts of the problem--recall that when you take an even root, both the positive and negative solutions are valid since both `(3)^4=81` and `(-3)^4=81`.

We are left with:

`x+5=+-3`

Split into two equations:

`x+5=3" "" "" "x+5=-3`

These give `x=-2` and `x=-8`, respectively.

Answer 2

Considering real and complex solutions, we obtain: `x=-8,-2,-5+-3i`

Explanation:

Take the square root of both sides. Recall to take the positive and negative roots.

`(x+5)^2=+-9`

Solving for the equation with `+9` by again taking the square root, we are left with the two equations:

`x+5=3" "" "" "x+5=-3`

Resulting in the two zeros `x=-2` and `x=-8`.

The other two zeros, which are complex, come from solving `(x+5)^2=-9`.

Taking the square root of both sides yields `x+5=+-3i`, so `x=-5+-3i`.