Question

If $\sqrt{2(x+24)}-\sqrt{x-7}\ge \sqrt{x+7},$ then $x\in$

• A. $[-25,25]$
• B. $(-17,-24)\cup [7,\infty )$
• C. $[7,\infty )$
• D. $[7,25]$

Answer 1

The correct option is D $[7,25]$

$\sqrt{2(x+24)}-\sqrt{x-7}\ge \sqrt{x+7}\cdots \left(1\right)$
For equation $\left(1\right)$ to be true,
$2(x+24)\ge 0\Rightarrow x\ge -24$
$x-7\ge 0\Rightarrow x\ge 7$
$x+7\ge 0\Rightarrow x\ge -7$
So, $x\in [7,\infty )\cdots \left(2\right)$
Now,
$\sqrt{2(x+24)}-\sqrt{x-7}\ge \sqrt{x+7}$
$\Rightarrow \sqrt{2(x+24)}\ge \sqrt{x-7}+\sqrt{x+7}$
Squaring both side, we get
$(\sqrt{2(x+24)}{)}^2\ge (\sqrt{x-7}+\sqrt{x+7}{)}^2$
$\Rightarrow 2(x+24)\ge x-7+x+7+2\sqrt{x^2-49}$
$\Rightarrow 24\ge \sqrt{x^2-49}$
Squaring both side again, we get
$576\ge x^2-49$
$\Rightarrow x^2-625\le 0$
$\Rightarrow (x-25)(x+25)\le 0\Rightarrow x\in [-25,25]\cdots \left(3\right)$
Taking intersection of equation $\left(2\right)$ and $\left(3\right),$ we get
$x\in [7,25]$