od Kosinus » Nedelja, 18. April 2021, 15:23
[dispmath]\sqrt{5-4x-x^2}\ge-2x-1[/dispmath][dispmath]P(x)=5-4x-x^2\\
Q(x)=-2x-1[/dispmath][dispmath]P(x)\ge 0\quad\land\quad Q(x)<0\\
5-4x-x^2\ge0\qquad\land\qquad-2x-1<0\\
x\in\Big[-5,1\Big]\qquad\land\qquad x\in\Big(\frac{-1}{2},\infty\Big)\\
\underline{\enspace x\in\Big(\frac{-1}{2},1\Big]\enspace}[/dispmath][dispmath]P(x)\ge Q^2(x)\quad\land\quad Q(x)\ge0\\
5-4x-x^2\ge(-2x-1)^2\qquad\land\qquad-2x-1\ge0\\
x\in\Big[-2,\frac{2}{5}\Big]\qquad\land\qquad x\in\Big(-\infty,\frac{-1}{2}\Big]\\
\underline{\enspace x\in\Big[-2,\frac{-1}{2}\Big]\enspace}[/dispmath]
[dispmath]x\in\Big(\frac{-1}{2},1\Big]\quad\lor\quad x\in\Big[-2,\frac{-1}{2}\Big][/dispmath][dispmath]\enclose{box}{x\in\Big[-2,1\Big]}[/dispmath]