Is the $\sqrt{\alpha+2y}$ in solving quartic equations with Ferrari method real?

Question

Assume I want to solve $$ u^4 + \alpha u^2 + \beta u + \gamma = 0 $$ (only real-valued solutions are needed) for real $\alpha, \beta, \gamma \in \mathbb R$, $\beta \ne 0$ and assume I have already found a $y\in \mathbb R$ such that $$ y^3 + \frac 5 2\alpha y^2 + (2\alpha^2 - \gamma) y + \bigg( \frac {\alpha^3} 2 - \frac {\alpha \gamma} 2 - \frac {\beta^2} 8 \bigg) = 0. $$ Does this imply $\alpha + 2 y \ge 0$?

I am asking because, as a next step, Wikipedia suggests to solve two quadratic equations in $u$, which contain $\sqrt{\alpha + 2 y}$.

If not, is there another method of solving the original quartic equation in $u$ with only real numbers involved?

Answer 1

Whether $\sqrt{\alpha+2y}$ is real is less important than recognizing that complex coefficients present no issue with solving quadratic equations. The quadratic formula may give a square root of a complex number, true; but that (unlike cube roots and higher prime-order roots) can be solved by algebra alone. See, for instance, this very informative answer on how to do it.

If the resolvent cubic given in the question has three real roots, then not all can give positive values of $\alpha+2y$ for positive $\alpha$. For the three real roots satisfy

$y_1+y_2+y_3=-(5/2)\alpha$

And thus

$(\alpha+2y_1)+(\alpha+2y_2)+(\alpha+2y_3)=3\alpha-5\alpha=-2\alpha$

which is negative for positive $\alpha$, from which st least one of the $\alpha+2y$ addends must also be negative.

Answer 2

I think if we have at least one $$ u \in \bigg( \mathbb R \backslash \bigg\{ \frac \beta {2(\alpha + 2y)} \bigg\} \bigg) \cup (i\cdot \mathbb R) $$ that solves the original quartic equation, the answer is yes, i.e. $\alpha + 2y \ge 0$. For the purpose of the question, where only real solutions $u$ are searched for, that statement is enough:

1. calculate all extreme points of the original quartic equation (solving cubic equations is possible with only real numbers involved)
2. If all extreme points are positive, there are no real solutions $u$. If at least one extreme point is non-positive, there are real solutions $u$. Then the above statement tells us that the quartic equation can be solved with only real numbers involved.

However, this strategy looks complicated, so further answers are appreciated!

Proof: Let us fix such a $u$. Then $u^2 \in \mathbb R$. Due to Wikipedia (equation (5)), our setting implies $$ \bigg( \sqrt{\alpha + 2y} u - \frac {\beta} {2\sqrt{\alpha + 2y}} \bigg)^2 = (u^2 + \alpha + y)^2 \in [0;\infty) $$ This implies $$ \sqrt{\alpha + 2y} u - \frac {\beta} {2\sqrt{\alpha + 2y}} \in \mathbb R $$ Now assume $\alpha + 2y < 0$. Then $$ \sqrt{\alpha + 2y}, \frac \beta {2\sqrt{\alpha + 2y}} \in i \cdot \mathbb R $$

If $u \in \mathbb R$, then we obtain $$ 0 = \sqrt{\alpha + 2y} u - \frac {\beta} {2\sqrt{\alpha + 2y}} \Rightarrow u = \frac \beta {2(\alpha + 2y)}, $$ a contradiction.

If $u \in i \cdot \mathbb R$, then we obtain $\sqrt{\alpha + 2y} u \in \mathbb R$ and hence $\beta = 0$, which is also a contradiction to the original assumption.

Answer 3

Actually, the same Wikipedia page in the chapter "Ferrari's solution in the special case of real coefficients" tells that at least for the largest real root $y$, the term $\alpha + 2y$ is non-negative.