Your answer is B, two complex roots and two real roots.
By factoring the original equation(which is a difference of two squares), you get:
[tex] x^{4}-81=0\\(x^{2}-9)(x^{2}+9)=0[/tex]
Because first root is also a difference of two squares, it factors into x - 3 and x +3, your two real roots.
When you factor the second root, the roots are x - 3i and x + 3i.
To prove this, let's multiply them back together:
[tex][(x-3)(x+3)][(x-3i)(x+3i)]=0\\\\(x^{2}+3x-3x-9)(x^{2}+3xi-3xi-9i^{2})=0\\\\(x^{2}+0x-9)(x^{2}+0xi-9(-1))=0\\\\(x^{2}-9)(x^{2}+9)=0\\\\x^{4}+9x^{2}-9x^{2}-81=0\\\\x^{4}-81=0[/tex]
We reached the equation we started with, so that means that the roots are:
x + 3,
x - 3,
x + 3i, and
x - 3i,
two of which are real and two are complex.
