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Find-x-R-x-3-2x-3-2x-5-4-x-1-




Question Number 157805 by Huy last updated on 28/Oct/21
Find x∈R:            x^3 −2x+3=(√(2x+5))+4(√(x−1))
$$\mathrm{Find}\:\mathrm{x}\in\mathbb{R}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{3}} −\mathrm{2x}+\mathrm{3}=\sqrt{\mathrm{2x}+\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{x}−\mathrm{1}} \\ $$
Answered by MJS_new last updated on 28/Oct/21
rhs defined for x≥1 and strictly increasing  lhs strictly increasing for x>((√6)/3)  ⇒ only 1 solution ≥1  ⇒ x=2 (obvious solution 8−4+3=(√9)+4(√1)=7)
$$\mathrm{rhs}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant\mathrm{1}\:\mathrm{and}\:\mathrm{strictly}\:\mathrm{increasing} \\ $$$$\mathrm{lhs}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{for}\:{x}>\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{only}\:\mathrm{1}\:\mathrm{solution}\:\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{x}=\mathrm{2}\:\left(\mathrm{obvious}\:\mathrm{solution}\:\mathrm{8}−\mathrm{4}+\mathrm{3}=\sqrt{\mathrm{9}}+\mathrm{4}\sqrt{\mathrm{1}}=\mathrm{7}\right) \\ $$

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