Question

Solve the following inequality:
$x^2⩾x(2+\sqrt{12-2x-x^2}).$

Answer 1

$x^2\ge x(2+\sqrt{12-2x-x^2})$

$x-2\ge \left(\sqrt{12-2x-x^2}\right)$

$\sqrt{12-2x-x^2}\le x-2$

$\sqrt{12-2x-x^2}$

$12-2x-x^2\ge 0$

$-12+2x+x^2\le 0$

$x^2+2x-12=0$

$x=\frac{-2+\sqrt{2^2-4\cdot 1(-12)}}{2\cdot 1}:\sqrt{13}-1$

$x=\frac{-2-\sqrt{2^2-4\cdot 1(-12)}}{2\cdot 1}:-1-\sqrt{13}$

$(x-(\sqrt{13}-1))(x-(-1-\sqrt{13}))\le 0$

$-1-\sqrt{13}\le x\le \sqrt{13}-1$

$\sqrt{12-2x-x^2}\le x-2$

$-x^2-2x+12\le {(x-2)}^2$

$-x^2-2x+12\le x^2-4x+4$

$-2x^2+2x+8\le 0$

$-x^2+x+4\le 0$

$x^2-x-4\ge 0$

$(x-\frac{1+\sqrt{17}}{2})(x-\frac{1-\sqrt{17}}{2})\ge 0$

$x\le \frac{1-\sqrt{17}}{2}orx\ge \frac{1+\sqrt{17}}{2}$

$\frac{1+\sqrt{17}}{2}\le x\le \sqrt{13}-1$

$\left[\begin{array}{cc}Solution:& \frac{1+\sqrt{17}}{2}\le x\le \sqrt{13}-1\\ Decimal:& 2.56155\cdots \le x\le 2.60555\dots \\ IntervalNotation:& [\frac{1+\sqrt{17}}{2},\sqrt{13}-1]\end{array}\right]$