Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$
Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$
From here I don't know how to proceed...
 A: Your method might work, but I couldn't really see how to proceed. A substitution that I am sure would work is as follows:
First expand your polynomial: $x^4-(m^2+2m+2)x^2+m^2+2m+1$.
Next, you can use your $a=x^2$ substitution: $a^2-(m^2+2m+2)a+m^2+2m+1$. Then, you use essentially the same $b$ substitution, but expanded: $b=m^2+2m+1$. Once you do this you get $a^2-(b+1)a+b$, you solve and get $a=1, a=b$ (verify yourself that they work). Then substitute back $a$ and $b$: $x^2=1, x^2=m^2+2m+1$, so your solutions for $x$ are: $x=1$, $x=-1$, $x=m+1$, $x=-m-1$. Therefore, since an extremely large or small value of $x$ will make the equation positive, your range of x-values are $(-\infty, -|m|]\cap[-1, 1]\cap[|m|, \infty)$. Note when $m=-2$, your range of x-values are $(-\infty, \infty)$
Edit: My steps are essentially your steps (expanded out), but I didn't use quadratic formula like on the fourth step, but factored it.
Using what you have:
$ a = x^2 $
$ b = (m + 1)^2$
$ a^2 - (1 + b)a + b \geq 0$
$a=b, a=1$ are solutions to the inequality, and solve for bounds
The rest is the same.
A: Here's how it's done in high school, using Vieta's relations and results on the sign of a quadratic polynomial:
First set $y=x^2$ and solve for $y$: you have the quadratic inequation
$$y^2-\bigl(1+(m+1)^2\bigr)y+(m+1)^2 \ge0,\qquad y\ge 0.$$
The discriminant is
$$\Delta=\bigl(1+(m+1)^2\bigr)^2-4(m+1)^2=\bigl(1-(m+1)^2\bigr)^2,$$
therefore the quadratic polynomial in $y$ has two roots $y_0$ and $y_1\;(y_0\le y_1)$.
Furthermore these roots are nonnegative, since $y_0+y_1=\bigl(1+(m+1)^2\bigr)^2>0\:$ and $\:y_0y_1=(+1)^2\ge 0$.
Conclusion; the original inequation is equivalent  to
$$0\le y\le y_0\;\text{ or }\;y\ge y_1\iff \begin{cases}-\sqrt{y_0\mathstrut}\le x\le \sqrt{y_0\mathstrut},\\
\quad\text{ or }\\ x\le -\sqrt{y_1\mathstrut},\enspace x\ge\sqrt{y_1\mathstrut}. \end{cases}$$
A: You made a mistake. From here
$$ a^2 - (1 + b)a + b \geq 0$$
you went here
$$ a=\frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
While the correct solution is
$$a=\frac{-(1 + b) + \sqrt{ (1 + b)^2-4b}}{2} =\frac{-(1 + b) + \sqrt{ (1 - b)^2}}{2}$$
$$a_1=1;\;a_2=b$$
$$a\le 1\lor a\ge b$$
that is
$$x^2\le 1 \lor x^2\ge (m+1)^2$$
or
$$-1\le x \le 1 \lor  x\ge| m+1|$$
