Question

Solve the following inequality:
$\frac{1-\sqrt{21-4x-x^2}}{x+1}⩾0.$

Answer 1

$\frac{1-\sqrt{21-4x-x^2}}{x+1}\ge 0$

at $x=-1,x+1=0$ so, $x\ne -1$

$1-\sqrt{21-4x-x^2}\ge 0$

$1\ge \sqrt{21-4x-x^2}$

Squaring on both the sides,

$1\ge (\sqrt{21-4x-x^2}{)}^2\ge 0$ (as 21-4x-x^{2} cannot be negative}

$1\ge 21-4x-x^2\ge 0$

$\Rightarrow -1\le x^2+4x-21\le 0$

$\Rightarrow -1\le (x-3)(x+7)\le 0$

Critical points are $3,-7$

Now for $(x-3)(x+7)\le 0$

$\Rightarrow x\in (-\infty ,-7]\cup [3,\infty )$