<s>[INST] Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$ 
Consider the bi-quadratic equation $E:x^4-5x^3+(\lambda+2)x^2-5x+1=0$ then, the real values of $\lambda$ so that $E$ has four different solutions is?

My attempts:
As $x=0$ is not a solution for any $\lambda$ hence we divide by $x^2$,\begin{align}x^2+\dfrac{1}{x^2}-5x-5\dfrac{1}{x}+\lambda+2&=0\\\bigg(x+\dfrac{1}{x}\bigg)^2-5\bigg(x+\dfrac{1}{x}\bigg)&=-\lambda\\t(t-5)&=-\lambda\end{align}
Here, $x+\dfrac{1}{x}=t$ cannot lie between $[-2,2]$.
Now we draw the graph of $-t(t-5)$ and $\lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+\lambda=0$, each of the roots of $Q$ gives two $x's$.
From GRAPH HERE, my solution is $\lambda\in(-\infty,-14]\cup[6,\frac{25}{4})$
But this wrong according to answer provided, please help.
 [/INST] Let $f(x)=-\frac{x^4-5x^3+2x^2-5x+1}{x^2}.$
We need to find all values of $\lambda,$ for which the equation
$$\lambda=f(x)$$ has four different roots.
Now, $$f'(x)=\frac{(2-x)(x-1)(2x-1)(x+1)}{x^3},$$ which says that
$$\lambda<f(-1)$$ or $$f(1)<\lambda<\min\left\{f\left(\frac{1}{2}\right),f(2)\right\},$$
which gives the answer:
$$(-\infty,-14)\cup(6,6.25).$$
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