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Question Number 117641 by bemath last updated on 13/Oct/20

$$\mathrm{find}\mathrm{the}\mathrm{solution}\sqrt{6-\mathrm{x}}>\mathrm{x}-4$$

Answered by Don08q last updated on 13/Oct/20

$$(6-x)>{(x-4)}^{2}or(6-x)<{(x-4)}^2$$ $$when6-x>x^2-8x+16$$ $$x^2-7x+10<0$$ $$(x-5)(x-2)<0$$ $$2<x<5$$ $$$$ $$when6-x<x^2-8x+16$$ $$x^2-7x+10>0$$ $$(x-5)(x-2)>0$$ $$x<2\cup x>5$$ $$Hence,\{2<x<5orx<2\cup x>5\}$$

Commented bybemath last updated on 13/Oct/20

$$\mathrm{your}\mathrm{solution}2<\mathrm{x}<5?$$ $$\mathrm{i}\mathrm{think}\mathrm{it}\mathrm{wrong}.\mathrm{let}\mathrm{substitute}$$ $$\mathrm{x}=-3\Rightarrow \sqrt{6-(-3)}>-3-4$$ $$\Rightarrow 3>-7\mathrm{it}\mathrm{true}$$

Answered by floor(10²Eta[1]) last updated on 13/Oct/20

$$\mathrm{x}<5$$

Answered by 1549442205PVT last updated on 13/Oct/20

$$\sqrt{6-\mathrm{x}}>\mathrm{x}-4\left(1\right)$$ $$\mathrm{We}\mathrm{need}\mathrm{the}\mathrm{condition}\mathrm{x}⩽6\mathrm{for}\mathrm{the}$$ $$\mathrm{root}\mathrm{is}\mathrm{defined}.\mathrm{Then}$$ $$\mathrm{i})\mathrm{For}\mathrm{x}<4\mathrm{then}\left(1\right)\mathrm{is}\mathrm{true}$$ $$\mathrm{ii})\mathrm{For}4⩽\mathrm{x}⩽6\left(2\right)\mathrm{then}$$ $$\left(1\right)\iff (6-\mathrm{x})>{(\mathrm{x}-4)}^2={\mathrm{x}}^2-8\mathrm{x}+16$$ $$\iff {\mathrm{x}}^2-7\mathrm{x}+10<0\iff (\mathrm{x}-2)(\mathrm{x}-5)<0$$ $$\iff 2<\mathrm{x}<5\left(3\right)$$ $$\mathrm{Combining}\left(2\right)\left(3\right)\mathrm{we}\mathrm{get}4⩽\mathrm{x}<5$$ $$.\mathrm{Combining}\mathrm{i})\mathrm{and}\mathrm{ii})\mathrm{we}\mathrm{get}\mathrm{solutions}$$ $$\mathrm{of}\mathrm{given}\mathrm{inequality}\mathrm{are}\mathrm{x}\in (-\mathrm{\infty },5)$$

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