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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:
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:
![[(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](https://tex.z-dn.net/?f=%5B%28x-3%29%28x%2B3%29%5D%5B%28x-3i%29%28x%2B3i%29%5D%3D0%5C%5C%5C%5C%28x%5E%7B2%7D%2B3x-3x-9%29%28x%5E%7B2%7D%2B3xi-3xi-9i%5E%7B2%7D%29%3D0%5C%5C%5C%5C%28x%5E%7B2%7D%2B0x-9%29%28x%5E%7B2%7D%2B0xi-9%28-1%29%29%3D0%5C%5C%5C%5C%28x%5E%7B2%7D-9%29%28x%5E%7B2%7D%2B9%29%3D0%5C%5C%5C%5Cx%5E%7B4%7D%2B9x%5E%7B2%7D-9x%5E%7B2%7D-81%3D0%5C%5C%5C%5Cx%5E%7B4%7D-81%3D0)
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.
By factoring the original equation(which is a difference of two squares), you get:
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:
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.