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Question Number 122322 by EngLewis last updated on 15/Nov/20

$$\mid x+1\mid <\mid x^2+2x+2\mid$$ $$solveforx$$

Answered by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{we}\mathrm{have}{\mathrm{x}}^2+2\mathrm{x}+2={(\mathrm{x}+1)}^2+1>0\Rightarrow \mid {\mathrm{x}}^2+2\mathrm{x}+2\mid ={\mathrm{x}}^2+2\mathrm{x}+2$$ $$\mathrm{e}\Rightarrow \mid \mathrm{x}+1\mid -{\mathrm{x}}^2-2\mathrm{x}-2<0\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mid \mathrm{x}+1\mid -{\mathrm{x}}^2-2\mathrm{x}-2$$ $$\mathrm{x}-\mathrm{\infty }-1+\mathrm{\infty }$$ $$\mid \mathrm{x}+1\mid -\mathrm{x}-10\mathrm{x}+1$$ $$\mathrm{f}\left(\mathrm{x}\right)-{\mathrm{x}}^2-3\mathrm{x}-3-{\mathrm{x}}^2-\mathrm{x}-1$$ $$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}-{\mathrm{x}}^2-3\mathrm{x}-3,\mathrm{x}⩽-1\\ -{\mathrm{x}}^2-\mathrm{x}-1,\mathrm{x}⩾-1\end{array}$$ $$\mathrm{case}1\mathrm{x}⩽-1\mathrm{f}\left(\mathrm{x}\right)<0\Rightarrow -{\mathrm{x}}^2-3\mathrm{x}-3<0\Rightarrow {\mathrm{x}}^2+3\mathrm{x}+3>0$$ $$\mathrm{\Delta }=9-12=-3\Rightarrow \mathrm{this}\mathrm{polynom}\mathrm{is}\mathrm{slways}>0\Rightarrow {\mathrm{S}}_1=]-\mathrm{\infty },-1]$$ $$\mathrm{case}2\mathrm{x}⩾1\mathrm{f}\left(\mathrm{x}\right)<0\Rightarrow -{\mathrm{x}}^2-\mathrm{x}-1<0\Rightarrow {\mathrm{x}}^2+\mathrm{x}+1>0$$ $$\mathrm{this}\mathrm{polynom}\mathrm{is}\mathrm{also}>0\Rightarrow {\mathrm{S}}_2=[-1,+\mathrm{\infty }[\mathrm{so}$$ $$\mathrm{S}=\cup {\mathrm{S}}_{\mathrm{i}}=\mathrm{R}$$

Answered by MJS_new last updated on 15/Nov/20

$$\mid x+1\mid <\mid {(x+1)}^2+1\mid$$ $$\mid x+1\mid <{(x+1)}^2+1$$ $$t=x+1$$ $$\mid t\mid <t^2+1$$ $$\left(1\right)t<0\Rightarrow -t<t^2+1\iff t^2+t+1>0\mathrm{always}\mathrm{true}$$ $$\left(2\right)t>0\Rightarrow t<t^2+1\iff t^2-t+1>0\mathrm{always}\mathrm{true}$$ $$\Rightarrow$$ $$\mathrm{solution}\mathrm{is}x\in \mathbb{R}$$

Answered by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{another}\mathrm{way}\mathrm{but}\mathrm{eazy}$$ $${({\mathrm{x}}^2+2\mathrm{x}+2)}^2-{(\mathrm{x}+1)}^2=({\mathrm{x}}^2+2\mathrm{x}+2-\mathrm{x}-1)({\mathrm{x}}^2+2\mathrm{x}+2+\mathrm{x}+1)$$ $$=({\mathrm{x}}^2+\mathrm{x}+1)({\mathrm{x}}^2+3\mathrm{x}+3)\mathrm{but}{\mathrm{x}}^2+\mathrm{x}+1>0\mathrm{\forall }\mathrm{x}\mathrm{due}\mathrm{yo}\mathrm{\Delta }<0$$ $${\mathrm{x}}^2+3\mathrm{x}+3>0\mathrm{\forall }\mathrm{x}\mathrm{due}\mathrm{to}\mathrm{\Delta }<0\Rightarrow \mathrm{\forall }\mathrm{x}\in \mathrm{R}{(\mathrm{x}+1)}^2<{({\mathrm{x}}^2+2\mathrm{x}+2)}^2\Rightarrow$$ $$\mid \mathrm{x}+1\mid <{\mathrm{x}}^2+2\mathrm{x}+2$$

Answered by TANMAY PANACEA last updated on 16/Nov/20

$$x^2+2x+2$$ $${(x+1)}^2+1$$ $${(x+1)}^{2}alwayspositive$$ $$so{(x+1)}^2+1alwaysgreaterthan(x+1)$$

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