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Question Number 115312 by naka3546 last updated on 25/Sep/20

Solve for x ∈ R that suitable on this inequality : (√(8−x^2 )) > x

$${Solve}\:\:{for}\:\:{x}\:\in\:\mathbb{R}\:\:{that}\:\:{suitable}\:\:{on}\:\:{this} \\ $$ $${inequality}\::\:\:\:\:\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\:>\:\:{x} \\ $$

Commented bybemath last updated on 25/Sep/20

intersection two graph we get 8−x^2 = x^2 ⇒ 8=2x^2 x = ± 2 ⇒ solution set is x< 2.

$${intersection}\:{two}\:{graph} \\ $$ $${we}\:{get}\:\mathrm{8}−{x}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \:\Rightarrow\:\mathrm{8}=\mathrm{2}{x}^{\mathrm{2}} \\ $$ $$\:{x}\:=\:\pm\:\mathrm{2}\:\Rightarrow\:{solution}\:{set}\:{is}\:\:{x}<\:\mathrm{2}. \\ $$

Commented byMJS_new last updated on 25/Sep/20

you are right, two graphs but f(x)=(√(8−x^2 )) is defined for −2(√2)≤x≤2(√2) and f(x)≥0. you are not allowed to square

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{two}\:\mathrm{graphs} \\ $$ $$\mathrm{but}\:{f}\left({x}\right)=\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $$\mathrm{and}\:{f}\left({x}\right)\geqslant\mathrm{0}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{not}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$

Answered by john santu last updated on 25/Sep/20

(1) 8−x^2 ≥0 ⇒((√8)+x)((√8)−x)≥0 −2(√2) ≤ x ≤ 2(√2) (2) 8−x^2 >x^2 →4−x^2 >0 (2+x)(2−x)>0 −2<x<2 solution set (1)∩(2)⇒−2<x<2

$$\left(\mathrm{1}\right)\:\mathrm{8}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\left(\sqrt{\mathrm{8}}+{x}\right)\left(\sqrt{\mathrm{8}}−{x}\right)\geqslant\mathrm{0} \\ $$ $$−\mathrm{2}\sqrt{\mathrm{2}}\:\leqslant\:{x}\:\leqslant\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $$\left(\mathrm{2}\right)\:\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \:\rightarrow\mathrm{4}−{x}^{\mathrm{2}} >\mathrm{0}\: \\ $$ $$\:\:\:\:\:\:\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)>\mathrm{0}\: \\ $$ $$\:\:\:\:\:\:−\mathrm{2}<{x}<\mathrm{2} \\ $$ $${solution}\:{set}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\Rightarrow−\mathrm{2}<{x}<\mathrm{2} \\ $$

Commented bynaka3546 last updated on 25/Sep/20

but why ? any other solution? x = −2(√2) is right answer.

$${but}\:\:{why}\:?\:\:{any}\:\:{other}\:\:\:{solution}? \\ $$ $${x}\:=\:−\mathrm{2}\sqrt{\mathrm{2}}\:\:{is}\:\:{right}\:\:{answer}. \\ $$

Commented byMJS_new last updated on 25/Sep/20

(2) is wrong for x<0

$$\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{for}\:{x}<\mathrm{0} \\ $$

Answered by MJS_new last updated on 25/Sep/20

(√(8−x^2 ))>x (√(8−x^2 )) defined for −2(√2)≤x≤2(√2) (√(8−x^2 ))≥0 ⇒ −2(√2)≤x≤0 is a part of the solution for x>0 we are allowed to square 8−x^2 >x^2 ∧x>0 ⇒ 0<x<2 ⇒ solution is −2(√2)≤x<2

$$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }>{x} \\ $$ $$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\geqslant\mathrm{0}\:\Rightarrow\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the} \\ $$ $$\:\:\:\:\:\mathrm{solution} \\ $$ $$\mathrm{for}\:{x}>\mathrm{0}\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$ $$\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \wedge{x}>\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{0}<{x}<\mathrm{2} \\ $$ $$ \\ $$ $$\Rightarrow\:\mathrm{solution}\:\mathrm{is}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}<\mathrm{2} \\ $$

Answered by mr W last updated on 25/Sep/20

8−x^2 ≥0 ⇒−2(√2)≤x≤2(√2) for −2(√2)≤x≤0: (√(8−x^2 ))>x is always true. for 0<x≤2(√2): 8−x^2 >x^2 ⇒x<2 answer: −2(√2)≤x<2

$$\mathrm{8}−{x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$ $$\Rightarrow−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $${for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{0}: \\ $$ $$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }>{x}\:{is}\:{always}\:{true}. \\ $$ $${for}\:\mathrm{0}<{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}}: \\ $$ $$\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \\ $$ $$\Rightarrow{x}<\mathrm{2} \\ $$ $$ \\ $$ $${answer}: \\ $$ $$−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}<\mathrm{2} \\ $$

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