The set of values of $a$ for which $x^4+4x^3+ax^2+4x+1=0$ has real roots

Question

The set of values of $a$ for which the equation $x^4+4x^3+ax^2+4x+1=0$ has real roots is given by $(-∞,m] U {n}$. Find the value of $√(n-m)$

I completed the square, and got the value of $m$ to be $6$, but I cannot find the value of y. Since I don't know the answer, I couldn't reverse engineer the problem :(

Answer 1

Note that$$x^4+4x^2+ax^3+4x+1=(x+1)^4-(6-a)x^2.$$So, if $a=6$, then your polynomial is $(x+1)^4$, which only has real roots. If $a>6$, then $6-a<0$ and therefore it has no real roots. If $a<6$, let $b=\sqrt{6-a}$. Then\begin{align}x^4+4x^3+ax^2+4x+1&=(x+1)^4-b^2x^2\\&=\bigl((x+1)^2-bx\bigr)\bigl((x+1)^2+bx\bigr)\\&=\bigl(x^2+(2-b)x+1\bigr)\bigl(x^2+(2+b)x+1\bigr).\end{align}Can you take it from here?

Answer 2

Note that $$x=0$$ is not a solution, so we can write $$x^2+\frac{1}{x^2}+4\left(x+\frac{1}{x}\right)+a=0$$ Now let $$t=x+\frac{1}{x}$$. Can you proceed?