Question

Consider the equation 256 = x4. How many real solutions of x exist?

Answer 1

$256 = x^{4}$

By subtracting 256 and moving terms, we get:
$x^{4} - 256 = 0$
$(x^{2^{2}} - 16^{2}) = 0$

This is a difference of two squares.
$(x^{2} - 4)(x^{2} + 4) = 0$
$\text{Since } x^{2} > 0\text{, there lies only two solutions.}$

We can see that for even powers, there will only lie two solutions, because they cannot lie less than zero. Any form of a quadratic will only have 2 solutions, because the remaining are all imaginary. For this example, we can see that, since the degree of the binomial is degree 4, there can only lie two solutions because the conjugate is also a solution.

For any quadratic or even degree functions, the conjugate solution has to be rejected for the equation to satisfy the laws of squares and square roots.

Thus, the remaining 2 solutions are imaginary, leaving only two real solutions.