(√(3−x))−(√(x+1))>(1/2)


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

$\sqrt{3 - x} - \sqrt{x + 1} > \frac{1}{2}$
Commented bykaivan.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 bykey 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 bykaivan.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 bykaivan.ahmadi last updated on 24/Jan/20

$a z i z d e l a m$
Commented byjohn santu last updated on 24/Jan/20
$(i) x≤3 ∧x≥−1 (ii) (√(3−x)) −(1/2)>(√(x+1)) squaring 3−x+(1/4)−(√(3−x)) >x+1 (9/4)−2x>(√(3−x)) , x<(9/8) squaring ⇒((81)/(16))−9x+4x^2 >3−x 4x^2 −8x+((33)/(16))>0 x^2 −2x+((33)/(64))>0 (x−1)^2 −((31)/(64))>0 x<1−((√(31))/8) ∨ x>1+((√(31))/8) now we get solution −1≤x≤3 ∧x<(9/8)∧{x<1−((√(31))/8) ∨x>1+((√(31))/8)} ∴ −1≤x<1−((√(31))/8)$
$(i) x ⩽ 3 \land x ⩾ - 1$ $(i i) \sqrt{3 - x} - \frac{1}{2} > \sqrt{x + 1}$ $s q u a r i n g$ $3 - x + \frac{1}{4} - \sqrt{3 - x} > x + 1$ $\frac{9}{4} - 2 x > \sqrt{3 - x}, x < \frac{9}{8}$ $s q u a r i n g \Rightarrow \frac{81}{16} - 9 x + 4 x^{2} > 3 - x$ $4 x^{2} - 8 x + \frac{33}{16} > 0$ $x^{2} - 2 x + \frac{33}{64} > 0$ ${(x - 1)}^{2} - \frac{31}{64} > 0$ $x < 1 - \frac{\sqrt{31}}{8} \lor x > 1 + \frac{\sqrt{31}}{8}$ $n o w w e g e t s o l u t i o n$ $- 1 ⩽ x ⩽ 3 \land x < \frac{9}{8} \land {x < 1 - \frac{\sqrt{31}}{8} \lor x > 1 + \frac{\sqrt{31}}{8}}$ $∴ - 1 ⩽ x < 1 - \frac{\sqrt{31}}{8}$
Commented byjohn 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 byjagoll last updated on 24/Jan/20

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

	$3 - x ⩾ 0 \land x + 1 ⩾ 0 \Rightarrow - 1 ⩽ x ⩽ 3 (i)$ $\sqrt{3 - x} > \sqrt{x + 1} + \frac{1}{2} \Rightarrow 3 - x > \frac{1}{4} + x + 1 + \sqrt{x + 1} \Rightarrow$ $\frac{7}{4} - 2 x > \sqrt{x + 1} (\sqrt{x + 1} ⩾ 0 \Rightarrow \frac{7}{4} - 2 x ⩾ 0 \Rightarrow x ⩽ \frac{7}{8} (ii))$ ${4 x}^{2} - 8 x + \frac{33}{16} > 0 \Rightarrow \frac{8 + \sqrt{31}}{8} < x \lor x < \frac{8 - \sqrt{31}}{8} (iii)$ $i \cap ii \cap iii = [- 1, \frac{8 - \sqrt{31}}{8}]$
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