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Question Number 83512 by jagoll last updated on 03/Mar/20

$$\mathrm{find}\mathrm{the}\mathrm{solution}$$ $$\frac{{4\mathrm{x}}^2}{{(1-\sqrt{2\mathrm{x}+1})}^2}<2\mathrm{x}+9$$

Answered by john santu last updated on 03/Mar/20

$$\left(1\right)2\mathrm{x}+9>0\Rightarrow \mathrm{x}>-\frac{9}{2}$$ $$\left(2\right)\mathrm{x}⩾-\frac{1}{2}$$ $$\left(3\right){4\mathrm{x}}^2<(2\mathrm{x}+9){(1-\sqrt{2\mathrm{x}+1})}^2$$ $$\mathrm{let}\sqrt{2\mathrm{x}+1}=\mathrm{t}\Rightarrow \mathrm{x}=\frac{{\mathrm{t}}^2-1}{2}$$ $${({\mathrm{t}}^2-1)}^2<{(\mathrm{t}-1)}^2({\mathrm{t}}^2+8)$$ $${(\mathrm{t}-1)}^2\{{(\mathrm{t}+1)}^2-({\mathrm{t}}^2+8)\}<0$$ $${(\mathrm{t}-1)}^2(2\mathrm{t}-7)<0$$ $$\Rightarrow \mathrm{t}<1\cup 1<\mathrm{t}<\frac{7}{2}$$ $$\Rightarrow \sqrt{2\mathrm{x}+1}<1\cup 1<\sqrt{2\mathrm{x}+1}<\frac{7}{2}$$ $$\Rightarrow \mathrm{x}<0\cup 0<\mathrm{x}<\frac{45}{8}$$ $$\mathrm{the}\mathrm{solution}\mathrm{is}\left(1\right)\wedge \left(2\right)\wedge \left(3\right)$$ $$-\frac{1}{2}⩽\mathrm{x}<0\cup 0<\mathrm{x}<\frac{45}{8}$$

Commented byjagoll last updated on 03/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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