vasorasy8

2022-07-10

$\left\{\begin{array}{l}{y}^{2}-3\ge 0\\ 16{y}^{4}-96{y}^{2}\ge 0\end{array}$
the solution for the first inequality is $y\le -\sqrt{3}$ or $y\ge \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6}\le y\le \sqrt{6}$. Then for my result the solution for the system is
Then for my result the solution for the system is or $\sqrt{3}\le y\le \sqrt{6}$

Kroatujon3

$\left\{\begin{array}{l}{y}^{2}-3\ge 0\\ 16{y}^{4}-96{y}^{2}\ge 0\end{array}$
To satisfy condition 1
${y}^{2}-3\ge 0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{2}\ge 3\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}-\sqrt{3}\ge y$
To satisfy condition 2:
$16{y}^{4}-96{y}^{2}\ge 0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}-\sqrt{6}\le y\le \sqrt{6}$
Then, to satisfy both conditions
$y\in \left(-\mathrm{\infty },-\sqrt{3}\right)\cap \left(\sqrt{3},+\mathrm{\infty }\right)\cap \left(-\sqrt{6},\sqrt{6}\right)=\left(-\sqrt{6},-\sqrt{3}\right)\cup \left(\sqrt{3},\sqrt{6}\right)$

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