Question

Solve the following inequality :

$\displaystyle{\dfrac{1-\sqrt{21-4x-x^2}}{x+1}\geq 0}$

Answer 1

-2√6 - 2 ≤ x < - 1  & 2√6 - 2 ≤ x ≤ 3)

Step-by-step explanation:

$\displaystyle{\dfrac{1-\sqrt{21-4x-x^2}}{x+1}\geq 0}$

$21 - 4x -  {x}^{2}  \geqslant 0 \\ 25 - 4 - 4x -  {x}^{2}  \geqslant 0 \\    {5}^{2}  -  {(x + 2)}^{2}  \geqslant 0 \\  - 5\leqslant  x + 2 \leqslant 5$

$- 7 \leqslant x \leqslant 3$

x + 1 can not be zero

x can not be -1

if x < -1 then denominator is -ve

so numerator should be -ve

$1 -  \sqrt{21 - 4x -  {x}^{2} }  \leqslant 0 \\  \sqrt{21 - 4x -  {x}^{2} }  \geqslant 1 \\  {5}^{2}  -  {(x + 2)}^{2}  \geqslant 1 \\  {(x + 2)}^{2}  \leqslant 24$

$-2 \sqrt{6}\leqslant x + 2 \leqslant  +  2 \sqrt{6} \\ x <  - 1 \\  - 2 \sqrt{6}  - 2 \leqslant x <  - 1 \: is \: one \: solution$

if x > -1 then denominator is +ve so numerator should be +ve as well

 1 - √(21 - 4x - x²) ≥ 0

=> √(21 - 4x - x²) ≤ 1

Squaring both sides

21 - 4x - x² ≤ 1

=> 25 - (x + 2)² ≤ 1

=> (x +  2)² ≥ 24

=>   -2√6   ≥ x + 2   ≥ 2√6

=> -2 - 2√6 ≥  x   ≥ 2√6 - 2

x > - 1

=> 2√6 - 2 , 3

Another solution (2√6 - 2 ≤ x ≤ 3)

Answer 2

Answer:

Hey mate please refer to the attachment

the final solution for X is

x→ [-2-2√6,-1)U[-2+2√6,3]

hope it helps u,