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Question Number 77864 by Pratah last updated on 11/Jan/20

Commented by Pratah last updated on 11/Jan/20

$solve the inequality$
	Answered by key of knowledge last updated on 11/Jan/20
	$x={x}+[x]⇒ x^2 −4(x−{x})+3≥0⇒x^2 −4x+3+4{x}≥0⇒ in x≤1∨3≤x⇒x^2 −4x+3 and ∀x,4{x}≥0⇒ (−∞,1]∩[3,+∞)⇒x^2 −4[x]+3≥0(i) 1≤x<2⇒x^2 −1≥0⇒x∈R−(−1,1) ⇒1≤x<2 (ii) 2≤x<3⇒x^2 −5≥0⇒x∈R−(−(√5),(√5)) ⇒(√5)≤x<3 (iii) i∪ii∪iii⇒x∈R−[2,(√5))$
	$x = {x} + [x] \Rightarrow$ $x^{2} - 4 (x - {x}) + 3 ⩾ 0 \Rightarrow x^{2} - 4 x + 3 + 4 {x} ⩾ 0 \Rightarrow$ $in x ⩽ 1 \lor 3 ⩽ x \Rightarrow x^{2} - 4 x + 3 and \forall x, 4 {x} ⩾ 0 \Rightarrow$ $(- \infty, 1] \cap [3, + \infty) \Rightarrow x^{2} - 4 [x] + 3 ⩾ 0 (i)$ $1 ⩽ x < 2 \Rightarrow x^{2} - 1 ⩾ 0 \Rightarrow x \in R - (- 1, 1)$ $\Rightarrow 1 ⩽ x < 2 (i i)$ $2 ⩽ x < 3 \Rightarrow x^{2} - 5 ⩾ 0 \Rightarrow x \in R - (- \sqrt{5}, \sqrt{5})$ $\Rightarrow \sqrt{5} ⩽ x < 3 (i i i)$ $i \cup i i \cup i i i \Rightarrow x \in R - [2, \sqrt{5})$
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