Question

If $\sqrt{x+6}<x-6$, then

• A. $x\in (6,\infty )$
• B. $x\in (10,\infty )$
• C. $x\in (6,10)$
• D. $x\in (-6,\infty )$

Answer 1

The correct option is B $x\in (10,\infty )$

For $\sqrt{x+6}<x-6$ to be defined
$x+6\ge 0$
$\Rightarrow x\ge -6\cdots \left(1\right)$
As we know that $\sqrt{x+6}\ge 0$
$\therefore x-6>0$
$\Rightarrow x>6\cdots \left(2\right)$

Now the inequality has only $+ive$ quantity
squaring both sides, we get
$x+6<(x-6{)}^2$
$\Rightarrow x+6<x^2-12x+36$
$\Rightarrow x^2-13x+30>0\Rightarrow (x-3)(x-10)>0\Rightarrow x\in (-\infty ,3)\cup (10,\infty )\cdots \left(3\right)$

Taking intersection of all the three equations , we get
$x\in (10,\infty )$