Question

The solution set of the inequality $\sqrt{x+14}<(x+2)$ is

• A. $x\in (1,\infty )$
• B. $x\in (-2,\infty )$
• C. $x\in (-\infty ,2)$
• D. $x\in (2,\infty )$

Answer 1

The correct option is D $x\in (2,\infty )$

$\sqrt{x+14}<(x+2)$ ... $\left(i\right)$
$x+14\ge 0$
$x\ge -14$
Also, $x+2>0$
$\Rightarrow x>-2$
On squaring (1), we get
$\Rightarrow x+14<(x+2{)}^2$
$\Rightarrow x^2+3x-10>0$
$\Rightarrow (x+5)(x-2)>0$
$\Rightarrow x\in (-\infty ,-5)\cup (2,\infty )$
Hence, the common solution to all the conditions is $(2,\infty )$.