3-x-x-1-gt-1-2-

Question Number 79306 by jagoll last updated on 24/Jan/20

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

Commented by kaivan.ahmadi last updated on 24/Jan/20

3 - x ⩾ 0 \Rightarrow x ⩽ 3

x + 1 ⩾ 0 \Rightarrow x ⩾ - 1

\Rightarrow - 1 ⩽ x ⩽ 3

3 - x + x - 1 - 2 \sqrt{(3 - x) (x + 1)} > \frac{1}{4} \Rightarrow

- 2 \sqrt{(3 - x) (x + 1)} > - \frac{7}{4} \Rightarrow \sqrt{(3 - x) (x + 1)} < \frac{7}{8} \Rightarrow

(3 - x) (x + 1) < \frac{49}{64} \Rightarrow 3 - x^{2} + 2 x + 3 < \frac{49}{64} \Rightarrow

x^{2} - 2 x - \frac{143}{64} > 0

Δ = 4 + \frac{143}{16} = \frac{207}{64}

x_{1, 2} = \frac{2 \pm \frac{\sqrt{207}}{8}}{2} = 1 \pm \frac{\sqrt{207}}{16} = 1 \pm \frac{3 \sqrt{23}}{16}

x \in [- 1, 1 - \frac{3 \sqrt{23}}{16}] \cup [1 + \frac{3 \sqrt{23}}{16}, 3]

Commented by key of knowledge last updated on 24/Jan/20

ahmadi aziz, for x = 3 :

\sqrt{3 - 3} - \sqrt{3 + 1} = 0 - 2 = - 2 ≯ \frac{1}{2}

Commented by kaivan.ahmadi last updated on 24/Jan/20

H i s i r, t h a n k y o u .

I d i d e r r, i n l i n e 4

3 - x + x + 1

Commented by kaivan.ahmadi last updated on 24/Jan/20

a z i z d e l a m

Commented by john santu last updated on 24/Jan/20

(i) x ⩽ 3 \land x ⩾ - 1

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

s q u a r i n g

3 - x + \frac{1}{4} - \sqrt{3 - x} > x + 1

\frac{9}{4} - 2 x > \sqrt{3 - x}, x < \frac{9}{8}

s q u a r i n g \Rightarrow \frac{81}{16} - 9 x + 4 x^{2} > 3 - x

4 x^{2} - 8 x + \frac{33}{16} > 0

x^{2} - 2 x + \frac{33}{64} > 0

{(x - 1)}^{2} - \frac{31}{64} > 0

x < 1 - \frac{\sqrt{31}}{8} \lor x > 1 + \frac{\sqrt{31}}{8}

n o w w e g e t s o l u t i o n

- 1 ⩽ x ⩽ 3 \land x < \frac{9}{8} \land {x < 1 - \frac{\sqrt{31}}{8} \lor x > 1 + \frac{\sqrt{31}}{8}}

∴ - 1 ⩽ x < 1 - \frac{\sqrt{31}}{8}

Commented by john santu last updated on 24/Jan/20

b y c h e k i n g

x \in [- 1, 1 - \frac{\sqrt{31}}{8}) \Rightarrow [- 1, 0.304)

p u t x = 0.2 \Rightarrow \sqrt{3 - 0.2} - \sqrt{1 + 0.2}

= 1.673 - 1.095 = 0.578 . t r u e

Commented by jagoll last updated on 24/Jan/20

thank you

Answered by key of knowledge last updated on 24/Jan/20

3 - x ⩾ 0 \land x + 1 ⩾ 0 \Rightarrow - 1 ⩽ x ⩽ 3 (i)

\sqrt{3 - x} > \sqrt{x + 1} + \frac{1}{2} \Rightarrow 3 - x > \frac{1}{4} + x + 1 + \sqrt{x + 1} \Rightarrow

\frac{7}{4} - 2 x > \sqrt{x + 1} (\sqrt{x + 1} ⩾ 0 \Rightarrow \frac{7}{4} - 2 x ⩾ 0 \Rightarrow x ⩽ \frac{7}{8} (ii))

{4 x}^{2} - 8 x + \frac{33}{16} > 0 \Rightarrow \frac{8 + \sqrt{31}}{8} < x \lor x < \frac{8 - \sqrt{31}}{8} (iii)

i \cap ii \cap iii = [- 1, \frac{8 - \sqrt{31}}{8}]