Question

please help in this question!! question is attached.​

Answer 1

Answer:

$2 \leqslant x < 2 \sqrt{ \frac{7}{3} }$

Step-by-step explanation:

$\sqrt{x + 3} > \sqrt{x - 1} + \sqrt{x - 2}$

Expression inside square root must be non negative, so

x+3 ≥ 0, x-1 ≥ 0 and x-2 ≥ 0

⇒ x ≥ -3, x ≥ 1 and x ≥ 2

⇒ x ≥ 2

Squaring both side of inequality, we get

$x + 3 > (x - 1) + (x - 2) + 2 \sqrt{x - 1} \sqrt{x - 2}$

$6 - x > 2 \sqrt{x - 1} \sqrt{x - 2} \geqslant 0$

So 6-x ≥ 0

⇒ x ≤ 6

Squaring the equation again, we get

$36 - 12x + {x}^{2} > 4(x - 1)(x - 2)$

$36 - 12x + {x}^{2} > 4( {x}^{2} - 3x + 2)$

$36 - 12x + {x}^{2} > 4 {x}^{2} - 12x + 8$

$3 {x}^{2} - (36 - 8) < 0$

${x}^{2} - \frac{28}{3} < 0$

$(x + \sqrt{ \frac{28}{3} })( x - \sqrt{ \frac{28}{3} }) < 0$

$- 2 \sqrt{ \frac{7}{3} } < x < 2 \sqrt{ \frac{7}{3} }$

x must also satisfy the inequalities

$2 \leqslant x \leqslant 6$

Hence

$2 \leqslant x < 2 \sqrt{ \frac{7}{3} }$