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Question Number 109699 by bemath last updated on 25/Aug/20

     x(x−1)^2  ≥ 12(x−1)

$$\:\:\:\:\:{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{12}\left({x}−\mathrm{1}\right) \\ $$

Answered by bemath last updated on 25/Aug/20

⇔ (x−1)(x(x−1)−12)≥0  ⇔(x−1)(x−4)(x+3)≥0  ⇔ −3≤x≤1 ∪ x≥4

$$\Leftrightarrow\:\left({x}−\mathrm{1}\right)\left({x}\left({x}−\mathrm{1}\right)−\mathrm{12}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−\mathrm{1}\right)\left({x}−\mathrm{4}\right)\left({x}+\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:−\mathrm{3}\leqslant{x}\leqslant\mathrm{1}\:\cup\:{x}\geqslant\mathrm{4} \\ $$

Commented by nimnim last updated on 25/Aug/20

Nice..

$$\mathrm{Nice}.. \\ $$

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