I was solving a question from the book Challenge and Thrill of Pre-College Mathematics by V Krishnamurthy and got stuck and confuse at the end.
The question is find the set of values of $x$ for which the given inequality is true. $\mid x^2+3x\mid \ge 2-x^2$
My attempts:
1) If $x^2+3x\ge 0$ $x^2+3x \ge 2-x^2 \implies 2x^2+3x-2\ge 0 \implies (x+2)(2x-1)\ge 0 \implies x\le -2, x\ge \frac 12$
2) If $x^2+3x\lt 0\implies x\le \frac{-2}{3}$
$x\le -2$ and $x\le \frac{-2}{3}\implies x\le -2$ So my answer is $x\le -2, x\ge \frac 12$ but when I checked $\frac{-2}{3}$ is also a solution and $x=-2 \implies x^2+3x=-2\lt 0$ but it should be positive. Please clarify this to me.
Thanks.
If $x^2+3x \ge 0$, then $x \ge 0$ or $x\le -3.$
Upon solving $x^2+3x \ge 2-x^2$ and combine with the condition above, the answer is $x \le -3$ or $x \ge \frac12$.
If $x^2 + 3x < 0$, then $-3<x<0$.
Upon solving $-x^2-3x \ge 2-x^2$ and combine with the condition above, the answer is $-3<x\le-\frac23$.
Combining the second and fifth point, the solution is $$\{x:x \le -3\} \cup \{x:-3<x \le -\frac23\} \cup\{x:x \ge \frac12\}= \{x:x \le -\frac23\} \cup\{x:x \ge \frac12\} $$