Finding the set of values so all roots of the equation are real numbers

Question

Find the set he set of all values ​​of $\alpha, α ∈ (-π/2 ,π/2)$ such that all solutions of $x^4-2x^2+ \tan(α)=0$ are real numbers.

This question was first asked here by a friend of mine, but he's new to the site and is not really aware of all the rules (I am helping him with that), so I am asking this here with more details. We are both high school students and this problem was adapted from an entrance exam we are studying for.

This is his work:

$0 \le (x^2-1)^2$

$(x^2-1)^2 + \tan(\alpha) = 1\implies \tan(\alpha) \le 1\implies \alpha \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$

This is my work:

$y^2 - 2y + \tan(\alpha) = 0$

$y = \dfrac{2 \pm \sqrt{4 - 4\tan(\alpha)}}{2}$

$4 - 4\tan(\alpha) \geq 0$

$\tan(\alpha) \geq 1 \implies \alpha \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$

Basically, we both got the same answers in different ways, but the problem says the answer is actually $\alpha \in \left[0, \dfrac{\pi}{4}\right]$, and it gives no explanation. We cannot find any mistakes in our work. Could anyone explain what's going on?

Answer 1

As stated in the comments, when $\tan(\alpha) < 0$, then $\sqrt{4 - 4 \tan(\alpha)} > 2$, so $y^2 - 2y + \tan(\alpha) = 0$ will have a positive and a negative solution, thus when we plug the negative solution back into $y = x^2$, $x$ will be a complex number. Hence $\tan(\alpha) \geq 0$ and so the set is $\left[0, \dfrac{\pi}{4}\right]$.

Answer 2

You could have avoided the problem computing the discriminant of the quartic equation (have a look here).

$$\Delta=256 \,(\tan (\alpha )-1)^2 \,\tan (\alpha )$$

Quoting Wikipedia :

"If $\Delta < 0$ then the equation has two distinct real roots and two complex conjugate non-real roots".

Then, the result since the sign of $\Delta$ is the sign of $\tan (\alpha )$.