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solve-for-x-y-z-C-2x-2-3x-13x-2-52x-40-6y-2-14x-x-2-220x-300-z-2-2z-12x-2-72x-132-exact-solutions-possible-in-all-cases-




Question Number 112053 by MJS_new last updated on 05/Sep/20
solve for x, y, z ∈C:  2x^2 −3x=(√(13x^2 −52x+40))  6y^2 −14x=(√(x^2 −220x+300))  z^2 −2z=(√(−12x^2 +72x−132))  [exact solutions possible in all cases]
$$\mathrm{solve}\:\mathrm{for}\:{x},\:{y},\:{z}\:\in\mathbb{C}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}=\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{52}{x}+\mathrm{40}} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} −\mathrm{14}{x}=\sqrt{{x}^{\mathrm{2}} −\mathrm{220}{x}+\mathrm{300}} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}=\sqrt{−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{72}{x}−\mathrm{132}} \\ $$$$\left[\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}\:\mathrm{in}\:\mathrm{all}\:\mathrm{cases}\right] \\ $$
Commented by MJS_new last updated on 06/Sep/20
beware of false solutions due to squaring!
$$\mathrm{beware}\:\mathrm{of}\:\mathrm{false}\:\mathrm{solutions}\:\mathrm{due}\:\mathrm{to}\:\mathrm{squaring}! \\ $$
Answered by john santu last updated on 06/Sep/20
(1) (2x^2 −3x)^2 =13x^2 −52x+40      4x^4 −12x^3 +9x^2 −13x^2 +52x−40=0     4x^4 −12x^3 −4x^2 +52x−40=0     x^4 −3x^3 −x^2 +13x−10=0   (x−1)(x^3 −2x^2 −x−10)=0  for x=1 ←rejected  → x^3 −2x^2 −x−10=0  factor of −10 is ±1, ±2, ±5, ±10
$$\left(\mathrm{1}\right)\:\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}\right)^{\mathrm{2}} =\mathrm{13}{x}^{\mathrm{2}} −\mathrm{52}{x}+\mathrm{40} \\ $$$$\:\:\:\:\mathrm{4}{x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} −\mathrm{13}{x}^{\mathrm{2}} +\mathrm{52}{x}−\mathrm{40}=\mathrm{0} \\ $$$$\:\:\:\mathrm{4}{x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{52}{x}−\mathrm{40}=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{13}{x}−\mathrm{10}=\mathrm{0} \\ $$$$\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{10}\right)=\mathrm{0} \\ $$$${for}\:{x}=\mathrm{1}\:\leftarrow{rejected} \\ $$$$\rightarrow\:{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{10}=\mathrm{0} \\ $$$${factor}\:{of}\:−\mathrm{10}\:{is}\:\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{5},\:\pm\mathrm{10} \\ $$$$ \\ $$$$ \\ $$

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