Answer
Answer C
Work Step by Step
Subtract $125$ from both sides:
$x^4-125x=0\\
\rightarrow x(x^3-125)=0\\\rightarrow x(x-5)(x^2+5x+25)=0$
The solutions for $(x^2+5x+25)=0$ will be:
$x=\frac{-b \pm \sqrt b^2-4ac}{2a}=\frac{-25\pm \sqrt 5^2-4(1)(25)}{2(1)}=\frac{-5\pm\sqrt -75}{2}$
These solutions are not real.
Hence, the only real solutions of the given equation are $x=0$ and $x=5$.
