<s>[INST] Quartic with 3 distinct roots Consider a quartic equation 
$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$
The value of $k$ so that given equation has three real and distinct solutions can be equal to -                                                                        i think that for any four degree to have 3 roots, its $f'(x)$ should have two zero and by the same way its $f''(x)$ should have one zero which means that the determinant of $f''(x)$ should be zero but this is wrong since the correct answer to this question is $\frac{13}{2}$.
 [/INST] Hint
Observe that $0$ is not a root of this equation and since three real roots exist, therefore fourth must be real as well and there must be a repeated root. 
Start with:
\begin{align*}
x^4 – kx^3 + 11x^2 – kx + 1 & = 0\\
x^2 – kx + 11 – \frac{k}{x} + \frac{1}{x^2} & = 0\\
\left(x+\frac{1}{x}\right)^2-k\left(x+\frac{1}{x}\right)+9 & =0\\
t^2-kt+9 & =0 && \left(\text{where } t=x+\frac{1}{x}\right)
\end{align*}
Now can you proceed from here to impose conditions for real roots?
</s>