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Question Number 138863 by bramlexs22 last updated on 19/Apr/21

$$\mid 2\mathrm{x}-1\mid ⩽\frac{4}{\sqrt{\mathrm{x}+2}}+1$$

Answered by lyubita last updated on 19/Apr/21

$$-2<x⩽2$$

Commented by bramlexs22 last updated on 19/Apr/21

$$\mathrm{by}\mathrm{Geogebra}$$

Commented by bramlexs22 last updated on 19/Apr/21

$$\mathrm{what}\mathrm{this}\mathrm{method}?$$

Answered by EDWIN88 last updated on 19/Apr/21

$$\mathrm{let}\sqrt{\mathrm{x}+2}=\mathrm{z};\mathrm{x}>-2$$$$\Rightarrow \mathrm{x}={\mathrm{z}}^2-2\wedge 2\mathrm{x}-1={2\mathrm{z}}^2-5$$$$\left(⋮\right)\mid {2\mathrm{z}}^2-5\mid ⩽\frac{4}{\mathrm{z}}+1$$$$\Rightarrow {\mathrm{z}}^2{({2\mathrm{z}}^2-5)}^2⩽{(\mathrm{z}+4)}^2$$$$\Rightarrow {\mathrm{z}}^2({4\mathrm{z}}^4-{20\mathrm{z}}^2+25)-{\mathrm{z}}^2-8\mathrm{z}-16⩽0$$$$\Rightarrow {4\mathrm{z}}^6-{20\mathrm{z}}^4+{24\mathrm{z}}^2-16⩽0$$$$\Rightarrow {\mathrm{z}}^6-{5\mathrm{z}}^4+{6\mathrm{z}}^2-4⩽0$$$$\mathrm{only}\mathrm{z}=2\mathrm{is}\mathrm{the}\mathrm{root}\mathrm{of}\mathrm{inequality}$$$$\Rightarrow \mathrm{z}⩽2;\mathrm{x}+2⩽4;\mathrm{x}⩽2$$$$\mathrm{the}\mathrm{solution}\mathrm{x}>-2\cap \mathrm{x}⩽2$$$$\Rightarrow \therefore -2<\mathrm{x}⩽2$$

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