Question

Решите неравенство $\sqrt{2x^{2} + 5x - 6} \ \textgreater \ \sqrt{-x-3}$

Answer 1

Ответ:

$\sqrt{2x^2+5x-6}>\sqrt{-x-3}\ \ \ \Leftrightarrow \ \ \ \left\{\begin{array}{l}2x^2+5x-6>-x-3\\-x-3\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}2x^2+6x-3>0\\-x\geq 3\end{array}\right\ \ \left\{\begin{array}{l}2\Big(x-\dfrac{-3-\sqrt{15}}{2}\Big)\Big(x-\dfrac{-3+\sqrt{15}}{2}\Big) >0\\x\leq -3\end{array}\right\\\\\\\left\{\begin{array}{l}x\in \Big(\infty \ ;\ \dfrac{-3-\sqrt{15}}{2}\ \Big)\cup \Big(\, \dfrac{-3+\sqrt{15}}{2}\ ;\ +\infty \, \Big)\\x\leq -3\end{array}\right$

$\dfrac{-3-\sqrt{15}}{2}\approx -3,4<-3\ \ \ ,\ \ \ \dfrac{-3+\sqrt{15}}{2}\approx 0,4\\\\\\Otvet:\ \ x\in \Big(-\infty \, ;\ \dfrac{-3-\sqrt{15}}{2}\ \Big)\ .$