2x-1-4-x-2-1-

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

∣ 2 x - 1 ∣ ⩽ \frac{4}{\sqrt{x + 2}} + 1

Answered by lyubita last updated on 19/Apr/21

- 2 < x ⩽ 2

Commented by bramlexs22 last updated on 19/Apr/21

by Geogebra

Commented by bramlexs22 last updated on 19/Apr/21

what this method ?

Answered by EDWIN88 last updated on 19/Apr/21

let \sqrt{x + 2} = z; x > - 2

\Rightarrow x = z^{2} - 2 \land 2 x - 1 = {2 z}^{2} - 5

(⋮) ∣ {2 z}^{2} - 5 ∣ ⩽ \frac{4}{z} + 1

\Rightarrow z^{2} {({2 z}^{2} - 5)}^{2} ⩽ {(z + 4)}^{2}

\Rightarrow z^{2} ({4 z}^{4} - {20 z}^{2} + 25) - z^{2} - 8 z - 16 ⩽ 0

\Rightarrow {4 z}^{6} - {20 z}^{4} + {24 z}^{2} - 16 ⩽ 0

\Rightarrow z^{6} - {5 z}^{4} + {6 z}^{2} - 4 ⩽ 0

only z = 2 is the root of inequality

\Rightarrow z ⩽ 2; x + 2 ⩽ 4; x ⩽ 2

the solution x > - 2 \cap x ⩽ 2

\Rightarrow ∴ - 2 < x ⩽ 2

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