Question

Solve the following inequality:
$2x+3<\sqrt{-2-3x-x^2}.$

Answer 1

$2x+3<\sqrt{-2-3x-x^2}$

$\sqrt{-2-3x-x^2}>2x+3$

$\sqrt{-2-3x-x^2}$

$-2-3x-x^2\ge 0$

$2+3x+x^2\le 0$

$(x+1)(x+2)\le 0$

$-2\le x\le -1$

$\sqrt{-2-3x-x^2}>2x+3$

$2x+3<0$

$x<-\frac{3}{2}$

$\sqrt{-2-3x-x^2}>2x+3$

$-x^2-3x-2>{(2x+3)}^2$

$-x^2-3x-2>4x^2+12x+9$

$-5x^2-15x-11>0$

$5x^2+15x+11<0$

$x=\frac{-15\pm \sqrt{{15}^2-4\cdot 5\cdot 11}}{2\cdot 5}:\frac{-15\pm \sqrt{5}}{10}$

$\frac{-15-\sqrt{5}}{10}<x<\frac{\sqrt{5}-15}{10}$

$x<-\frac{3}{2}or(x\ge -\frac{3}{2}and\frac{-15-\sqrt{5}}{10}<x<\frac{\sqrt{5}-15}{10})$

$x<-\frac{3}{2}or-\frac{3}{2}\le x<\frac{\sqrt{5}-15}{10}$

$-2\le x<\frac{\sqrt{5}-15}{10}$

$\left[\begin{array}{cc}Solution:& -2\le x<\frac{\sqrt{5}-15}{10}\\ Decimal:& -2\le x<-1.27639\dots \\ IntervalNotation:& [-2,\frac{\sqrt{5}-15}{10})\end{array}\right]$