Question

Find the values of $x$ satisfying: $2-3x-2x^2\ge 0$.

• A. $-2<x<\frac{1}{2}$
• B. $-2\le x\le \frac{1}{2}$
• C. $-\infty \le x\le -2$
• D. $\frac{1}{2}\le x\le \infty$

Answer 1

The correct option is B $-2\le x\le \frac{1}{2}$

We have, $2-3x-2x^2\ge 0$
$\Rightarrow 2x^2+3x-2\le 0$
$\Rightarrow (x+2)(2x-1)\le 0$
$\Rightarrow -2\le x\le \frac{1}{2}$