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]


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Question Number 112053 by MJS_new last updated on 05/Sep/20

$solve for x, y, z \in C :$ $2 x^{2} - 3 x = \sqrt{13 x^{2} - 52 x + 40}$ $6 y^{2} - 14 x = \sqrt{x^{2} - 220 x + 300}$ $z^{2} - 2 z = \sqrt{- 12 x^{2} + 72 x - 132}$ $[exact solutions possible in all cases]$
Commented by MJS_new last updated on 06/Sep/20

$beware of false solutions due to squaring!$
	Answered by john santu last updated on 06/Sep/20

	$(1) {(2 x^{2} - 3 x)}^{2} = 13 x^{2} - 52 x + 40$ $4 x^{4} - 12 x^{3} + 9 x^{2} - 13 x^{2} + 52 x - 40 = 0$ $4 x^{4} - 12 x^{3} - 4 x^{2} + 52 x - 40 = 0$ $x^{4} - 3 x^{3} - x^{2} + 13 x - 10 = 0$ $(x - 1) (x^{3} - 2 x^{2} - x - 10) = 0$ $f o r x = 1 \leftarrow r e j e c t e d$ $\to x^{3} - 2 x^{2} - x - 10 = 0$ $f a c t o r o f - 10 i s \pm 1, \pm 2, \pm 5, \pm 10$
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