Question

The solution set of the inequality $|3x-2|>|x+4|$ is

• A. $(-\infty ,-3)\cup (\frac{1}{2},\infty )$
• B. $(-3,\infty )$
• C. $(-\frac{1}{2},3)$
• D. $(-\infty ,-\frac{1}{2})\cup (3,\infty )$

Answer 1

The correct option is D $(-\infty ,-\frac{1}{2})\cup (3,\infty )$

$|3x-2|>|x+4|$
$\Rightarrow (3x-2{)}^2>(x+4{)}^2$
$\Rightarrow 9x^2-12x+4>x^2+8x+16$
$\Rightarrow 8x^2-20x-12>0$
$\Rightarrow 2x^2-5x-3>0$
$\Rightarrow (2x+1)(x-3)>0$
$\Rightarrow x\in (-\infty ,-\frac{1}{2})\cup (3,\infty )$