Question

If $x$ is real and $5x^2+2x+9>3x^2+10x+7$, then $x$ lies in the interval

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

Answer 1

Answer: (B)

$(-\infty ,2-\sqrt{3})\cup (2+\sqrt{3},\infty )$

$2x^2-8x+2>0$
$[x-2+\sqrt{3}][x-2-\sqrt{3}]>0$
Option B.