- #1
Siron
- 150
- 0
Hello!
Suppose a quartic polynomial with real roots where all its roots lie in the interval
$$x \in \left[\frac{5}{8}a-\frac{3}{8}\sqrt{15a^2-16b}, \frac{5}{8}a+\frac{3}{8}\sqrt{15a^2-16b}\right]$$
where $a \in \mathbb{R}$ and $b<(15/16)a^2$. Is there a way to check that at least one of those roots satisfies the inequality:
$$\frac{5ax-4x^2-b}{15ax-20x^2-3b}<0.$$
I solved the above inequality with wolfram alpha which displays the following solutions (after summarizing cases and taking $b<(15/16)a^2$ into account):
$$x \in \left[\frac{5a}{8}-\frac{1}{8}\sqrt{25a^2-16b}, \frac{3a}{8}-\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}\right]$$
or
$$x \in \left[\frac{3a}{8}+\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}, \frac{5a}{8}+\frac{1}{8}\sqrt{25a^2-16b} \right].$$
So ... in fact I think I only need to check wether the interval for my roots lies in one of the above intervals. However, I think this will result in too strong restrains on $b$. Is there a better alternative to solve the question?
Many thanks!
Kind regards,
Siron
Suppose a quartic polynomial with real roots where all its roots lie in the interval
$$x \in \left[\frac{5}{8}a-\frac{3}{8}\sqrt{15a^2-16b}, \frac{5}{8}a+\frac{3}{8}\sqrt{15a^2-16b}\right]$$
where $a \in \mathbb{R}$ and $b<(15/16)a^2$. Is there a way to check that at least one of those roots satisfies the inequality:
$$\frac{5ax-4x^2-b}{15ax-20x^2-3b}<0.$$
I solved the above inequality with wolfram alpha which displays the following solutions (after summarizing cases and taking $b<(15/16)a^2$ into account):
$$x \in \left[\frac{5a}{8}-\frac{1}{8}\sqrt{25a^2-16b}, \frac{3a}{8}-\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}\right]$$
or
$$x \in \left[\frac{3a}{8}+\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}, \frac{5a}{8}+\frac{1}{8}\sqrt{25a^2-16b} \right].$$
So ... in fact I think I only need to check wether the interval for my roots lies in one of the above intervals. However, I think this will result in too strong restrains on $b$. Is there a better alternative to solve the question?
Many thanks!
Kind regards,
Siron