Find the solution set of inequality (√(3x+4)) ≥ x−2


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Question Number 153070 by bobhans last updated on 04/Sep/21

$F i n d t h e s o l u t i o n s e t o f i n e q u a l i t y$ $\sqrt{3 x + 4} ⩾ x - 2$
	Answered by Rasheed.Sindhi last updated on 04/Sep/21

	$\sqrt{3 x + 4} ⩾ x - 2$ ${(\sqrt{3 x + 4} ⩾ x - 2)}^{2}$ $3 x + 4 ⩾ x^{2} - 4 x + 4$ $x^{2} - 7 x ⩽ 0$ $x (x - 7) ⩽ 0$ $(x ⩽ 0 \land x - 7 ⩾ 0) \lor (x ⩾ 0 \land x - 7 ⩽ 0)$ $(x ⩽ 0 \land x ⩾ 7) \lor (x ⩾ 0 \land x ⩽ 7)$ $0 ⩽ x ⩽ 7$
	Commented by MJS_new last updated on 04/Sep/21

	$\sqrt{3 x + 4} defined for x ⩾ - \frac{4}{3}$ $\sqrt{3 x + 4} = x - 2 \Rightarrow only one solution (squaring$ $leads to false solutions) : x = 7$ $\Rightarrow$ $the inequality holds for - \frac{4}{3} ⩽ x ⩽ 7$
	Commented by Rasheed.Sindhi last updated on 04/Sep/21

	$THANKS Sir! I think Ineed more$ $study regarding the topic .$
	Answered by john_santu last updated on 04/Sep/21
	${ ((x≥−(4/3); when x−2<0 we get solution −(4/3)≤x<2)),((when x−2≥0;x≥2⇒squaring both sides)) :} ⇒3x+4 ≥ x^2 −4x+4 ⇒x^2 −7x≤0 ; 0≤x≤7 we get solution 2≤x≤7 Therefore the solution set is ⇒x ∈ [−(4/3), 7 ]$
	${\begin{cases} x ⩾ - \frac{4}{3}; w h e n x - 2 < 0 w e g e t s o l u t i o n - \frac{4}{3} ⩽ x < 2 \\ w h e n x - 2 ⩾ 0; x ⩾ 2 \Rightarrow s q u a r i n g b o t h s i d e s \end{cases}$ $\Rightarrow 3 x + 4 ⩾ x^{2} - 4 x + 4$ $\Rightarrow x^{2} - 7 x ⩽ 0; 0 ⩽ x ⩽ 7 w e g e t s o l u t i o n 2 ⩽ x ⩽ 7$ $T h e r e f o r e t h e s o l u t i o n s e t i s$ $\Rightarrow x \in [- \frac{4}{3}, 7]$
	Answered by talminator2856791 last updated on 04/Sep/21

	$0 ⩾ x - 2 - \sqrt{3 x + 4}$ $0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})$ $0 ⩾ {(x - 2)}^{2} - (3 x + 4)$ $0 ⩾ x^{2} - 4 x + 4 - 3 x - 4$ $0 ⩾ x^{2} - 7 x$ $0 ⩾ x (x - 7)$ $0 ⩽ x ⩽ 7$
	Commented by MJS_new last updated on 04/Sep/21

	$0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})$ $if (x - 2) + \sqrt{3 x + 4} ⩾ 0 \Leftrightarrow x ⩾ 0$ $0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})$ $if (x - 2) + \sqrt{3 x + 4} < 0 \Leftrightarrow - \frac{4}{3} ⩽ x < 0$ $anyway you cannot solve it this way$ $y_{1} = x - 2 is a straight line$ $y_{2} = \sqrt{3 x + 4} is a half parabola$ $how ever you reach 3 x + 4 you get false solutions$ $\sqrt{3 x + 4} is defined for x ⩾ - \frac{4}{3}$ $x - 2 = \sqrt{3 x + 4} has only 1 solution : x = 7$ ${it}^{'} s easy to show that$ $x - 2 < \sqrt{3 x + 4} for - \frac{4}{3} ⩽ x < 0$ $and from both x - 2 and \sqrt{3 x + 4} strictly$ $increasing it follows that x - 2 > \sqrt{3 x + 4} for$ $x > 7$ $\Rightarrow solution is - \frac{4}{3} ⩽ x ⩽ 7$
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