Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$

Question

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.

Answer 1

Hint: Draw a graph $$f(x)= {-x^4+5x^3-2x^2+5x-1\over x^2}$$ and see for which $\lambda$ the line $y= \lambda$ cuts the graph of $f$ four times.

Answer 2

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: $$(-\infty,-14)\cup(6,6.25).$$