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Question Number 8234 by sandy_suhendra last updated on 03/Oct/16

$$\mathrm{Question}:\mathrm{figure}\mathrm{x}\mathrm{for}$$ $$\sqrt{\mathrm{x}-4}>6-\mathrm{x}$$ $$\mathrm{my}\mathrm{answer}:$$ $$\left(1\right)\mathrm{x}-4>{(6-\mathrm{x})}^2$$ $$(\mathrm{x}-5)(\mathrm{x}-8)<0$$ $$5<\mathrm{x}<8$$ $$$$ $$\left(2\right)\mathrm{x}-4⩾0$$ $$\mathrm{x}⩾4$$ $$\mathrm{so}\mathrm{I}\mathrm{have}\mathrm{for}\mathrm{x}\Rightarrow 5<\mathrm{x}<8$$ $${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{wrong}\mathrm{with}\mathrm{this}\mathrm{answer},\mathrm{please}\mathrm{help}\mathrm{me}$$ $$\mathrm{because}\mathrm{if}\mathrm{x}=9\Rightarrow \sqrt{9-4}>6-9,{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{true}$$

Commented byRasheed Soomro last updated on 03/Oct/16

$$\mathrm{when}\mathrm{you}\mathrm{square}\mathrm{to}\mathrm{both}\mathrm{sides}\mathrm{of}$$ $$\mathrm{inequality}/\mathrm{equation},\mathrm{the}\mathrm{result}\mathrm{is}$$ $$\mathrm{not}\mathrm{necessarily}\mathrm{completely}\mathrm{equivalent}$$ $$\mathrm{to}\mathrm{the}\mathrm{original}.$$ $$\mathrm{Your}\mathrm{solution}5<\mathrm{x}<8\mathrm{is}\mathrm{actually}\mathrm{the}$$ $$\mathrm{solution}\mathrm{of}\mathrm{x}-4>{(6-\mathrm{x})}^2\mathrm{which}\mathrm{may}$$ $$\mathrm{not}\mathrm{satisfy}\mathrm{the}\mathrm{original}\mathrm{inequation}$$ $$\mathrm{If}\mathrm{x}=9\Rightarrow 9-4>{(6-9)}^2\Rightarrow 5>9\mathrm{which}\mathrm{is}\mathrm{false}.$$

Commented byYozzias last updated on 03/Oct/16

$$\mathrm{Let}\mathrm{u}=\sqrt{\mathrm{x}-4}⩾0\mathrm{for}\mathrm{x}⩾4\Rightarrow \mathrm{x}={\mathrm{u}}^2+4$$ $$\therefore \mathrm{u}>6-{\mathrm{u}}^2-4$$ $${\mathrm{u}}^2+\mathrm{u}-2>0$$ $$(\mathrm{u}+2)(\mathrm{u}-1)>0$$ $$\Rightarrow \mathrm{u}>1\mathrm{or}\mathrm{u}<-2$$ $$\mathrm{But},\mathrm{u}⩾0\Rightarrow \mathrm{u}<-2\mathrm{is}\mathrm{not}\mathrm{possible}.$$ $$\therefore \mathrm{u}>1\Rightarrow \sqrt{\mathrm{x}-4}>1\Rightarrow \mathrm{x}-4>1\Rightarrow \mathrm{x}>5$$

Commented bysou1618 last updated on 04/Oct/16

$$ithink....$$ $$X^2<Y^2$$ $$\iff \mid X\mid <\mid Y\mid$$ $$youranswer$$ $$x-4>{(6-x)}^{2}means$$ $$\iff \sqrt{x-4}>\mid 6-x\mid$$ $$\iff +\sqrt{x-4}>6-x>-\sqrt{x-4}$$ $$isnotequalto{}^{\prime }{Question}^{\prime }$$ $$$$ $$youshould....$$ $$\left(i\right)if6-x⩾0(6⩾x)$$ $$\sqrt{x-4}>6-x⩾0$$ $$\iff x-4>{(6-x)}^2$$ $$\iff 0>x^2-13x+40$$ $$\iff 5<x<8$$ $$so$$ $$5<x⩽6$$ $$$$ $$\left(ii\right)if6-x⩽0(6⩽x)$$ $$\sqrt{x-4}⩾0⩾6-x$$ $$\iff x:allofthereal$$ $$so$$ $$6⩽x$$ $$$$ $$\left(i\right),\left(ii\right)\Rightarrow$$ $$5<x$$

Commented bysandy_suhendra last updated on 04/Oct/16

$${\mathrm{thank}}^{\prime }\mathrm{s}\mathrm{for}\mathrm{all}\mathrm{of}\mathrm{your}\mathrm{answers},\mathrm{I}\mathrm{really}\mathrm{appreciate}$$

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