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Question Number 153070 by bobhans last updated on 04/Sep/21
Find the solution set of inequality (√(3x+4)) ≥ x−2
$$\:\:\:{Find}\:{the}\:{solution}\:{set}\:{of}\:{inequality} \\ $$$$\:\:\:\:\:\:\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\: \\ $$
Answered by Rasheed.Sindhi last updated on 04/Sep/21
(√(3x+4)) ≥ x−2 ( (√(3x+4)) ≥ x−2 )^2 3x+4≥x^2 −4x+4 x^2 −7x≤0 x(x−7)≤0 (x≤0 ∧ x−7≥0) ∨ (x≥0∧x−7≤0) (x≤0 ∧ x≥7) ∨ (x≥0∧x≤7) 0≤x≤7
$$\:\:\:\:\:\:\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\: \\ $$$$\:\:\:\:\:\:\:\left(\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\geqslant\:{x}−\mathrm{2}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{3}{x}+\mathrm{4}\geqslant{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{7}{x}\leqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}\left({x}−\mathrm{7}\right)\leqslant\mathrm{0} \\ $$$$\:\left({x}\leqslant\mathrm{0}\:\wedge\:{x}−\mathrm{7}\geqslant\mathrm{0}\right)\:\vee\:\left({x}\geqslant\mathrm{0}\wedge{x}−\mathrm{7}\leqslant\mathrm{0}\right) \\ $$$$\:\left({x}\leqslant\mathrm{0}\:\wedge\:{x}\geqslant\mathrm{7}\right)\:\vee\:\left({x}\geqslant\mathrm{0}\wedge{x}\leqslant\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{7} \\ $$
Commented by MJS_new last updated on 04/Sep/21
(√(3x+4)) defined for x≥−(4/3) (√(3x+4))=x−2 ⇒ only one solution (squaring leads to false solutions): x=7 ⇒ the inequality holds for −(4/3)≤x≤7
$$\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\sqrt{\mathrm{3}{x}+\mathrm{4}}={x}−\mathrm{2}\:\Rightarrow\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\left(\mathrm{squaring}\right. \\ $$$$\left.\mathrm{leads}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions}\right):\:{x}=\mathrm{7} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{holds}\:\mathrm{for}\:−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}\leqslant\mathrm{7} \\ $$
Commented by Rasheed.Sindhi last updated on 04/Sep/21
THANKS Sir! I think Ineed more study regarding the topic.
$$\mathrm{THANKS}\:\mathrm{Sir}!\:\mathrm{I}\:\mathrm{think}\:\mathrm{Ineed}\:\mathrm{more} \\ $$$$\mathrm{study}\:\mathrm{regarding}\:\mathrm{the}\:\mathrm{topic}. \\ $$
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 ]
$$\:\begin{cases}{{x}\geqslant−\frac{\mathrm{4}}{\mathrm{3}};\:{when}\:{x}−\mathrm{2}<\mathrm{0}\:{we}\:{get}\:{solution}\:−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}<\mathrm{2}}\\{{when}\:{x}−\mathrm{2}\geqslant\mathrm{0};{x}\geqslant\mathrm{2}\Rightarrow{squaring}\:{both}\:{sides}}\end{cases} \\ $$$$\Rightarrow\mathrm{3}{x}+\mathrm{4}\:\geqslant\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{7}{x}\leqslant\mathrm{0}\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{7}\:{we}\:{get}\:{solution}\:\mathrm{2}\leqslant{x}\leqslant\mathrm{7} \\ $$$${Therefore}\:{the}\:{solution}\:{set}\:{is}\: \\ $$$$\:\Rightarrow{x}\:\in\:\left[−\frac{\mathrm{4}}{\mathrm{3}},\:\mathrm{7}\:\right] \\ $$
Answered by talminator2856791 last updated on 04/Sep/21
0 ≥ x−2 − (√(3x+4)) 0 ≥ ((x−2)−(√(3x+4)))((x−2)+(√(3x+4))) 0 ≥ (x−2)^2 −(3x+4) 0 ≥ x^2 −4x+4 − 3x−4 0 ≥ x^2 − 7x 0 ≥ x(x−7) 0 ≤ x ≤ 7
$$\: \\ $$$$\:\mathrm{0}\:\geqslant\:{x}−\mathrm{2}\:−\:\sqrt{\mathrm{3}{x}+\mathrm{4}} \\ $$$$\:\mathrm{0}\:\geqslant\:\left(\left({x}−\mathrm{2}\right)−\sqrt{\mathrm{3}{x}+\mathrm{4}}\right)\left(\left({x}−\mathrm{2}\right)+\sqrt{\mathrm{3}{x}+\mathrm{4}}\right) \\ $$$$\:\mathrm{0}\:\geqslant\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{3}{x}+\mathrm{4}\right) \\ $$$$\:\mathrm{0}\:\geqslant\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\:−\:\mathrm{3}{x}−\mathrm{4} \\ $$$$\:\mathrm{0}\:\geqslant\:{x}^{\mathrm{2}} \:−\:\mathrm{7}{x}\: \\ $$$$\:\mathrm{0}\:\geqslant\:{x}\left({x}−\mathrm{7}\right) \\ $$$$\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{7} \\ $$$$\: \\ $$
Commented by MJS_new last updated on 04/Sep/21
0≥((x−2)−(√(3x+4)))((x−2)+(√(3x+4))) if (x−2)+(√(3x+4))≥0 ⇔ x≥0 0≥((x−2)−(√(3x+4)))((x−2)+(√(3x+4))) if (x−2)+(√(3x+4))<0 ⇔ −(4/3)≤x<0 anyway you cannot solve it this way y_1 =x−2 is a straight line y_2 =(√(3x+4)) is a half parabola how ever you reach 3x+4 you get false solutions (√(3x+4)) is defined for x≥−(4/3) x−2=(√(3x+4)) has only 1 solution: x=7 it′s easy to show that x−2<(√(3x+4)) for −(4/3)≤x<0 and from both x−2 and (√(3x+4)) strictly increasing it follows that x−2>(√(3x+4)) for x>7 ⇒ solution is −(4/3)≤x≤7
$$\mathrm{0}\geqslant\left(\left({x}−\mathrm{2}\right)−\sqrt{\mathrm{3}{x}+\mathrm{4}}\right)\left(\left({x}−\mathrm{2}\right)+\sqrt{\mathrm{3}{x}+\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\mathrm{if}\:\left({x}−\mathrm{2}\right)+\sqrt{\mathrm{3}{x}+\mathrm{4}}\geqslant\mathrm{0}\:\Leftrightarrow\:{x}\geqslant\mathrm{0} \\ $$$$\mathrm{0}\geqslant\left(\left({x}−\mathrm{2}\right)−\sqrt{\mathrm{3}{x}+\mathrm{4}}\right)\left(\left({x}−\mathrm{2}\right)+\sqrt{\mathrm{3}{x}+\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\mathrm{if}\:\left({x}−\mathrm{2}\right)+\sqrt{\mathrm{3}{x}+\mathrm{4}}<\mathrm{0}\:\Leftrightarrow\:−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}<\mathrm{0} \\ $$$$ \\ $$$$\mathrm{anyway}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way} \\ $$$${y}_{\mathrm{1}} ={x}−\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$${y}_{\mathrm{2}} =\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{half}\:\mathrm{parabola} \\ $$$$\mathrm{how}\:\mathrm{ever}\:\mathrm{you}\:\mathrm{reach}\:\mathrm{3}{x}+\mathrm{4}\:\mathrm{you}\:\mathrm{get}\:\mathrm{false}\:\mathrm{solutions} \\ $$$$\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}−\mathrm{2}=\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{has}\:\mathrm{only}\:\mathrm{1}\:\mathrm{solution}:\:{x}=\mathrm{7} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$${x}−\mathrm{2}<\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{for}\:−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}<\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{from}\:\mathrm{both}\:{x}−\mathrm{2}\:\mathrm{and}\:\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{strictly} \\ $$$$\mathrm{increasing}\:\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:{x}−\mathrm{2}>\sqrt{\mathrm{3}{x}+\mathrm{4}}\:\mathrm{for} \\ $$$${x}>\mathrm{7} \\ $$$$\Rightarrow\:\mathrm{solution}\:\mathrm{is}\:−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}\leqslant\mathrm{7} \\ $$

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