Solve-for-real-numbers-2x-2-3y-2-z-2-7-x-2-y-2-z-2-2-z-x-y-

Question Number 171549 by Shrinava last updated on 17/Jun/22

Solve for real numbers :

{\begin{cases} {2 x}^{2} + {3 y}^{2} + z^{2} = 7 \\ x^{2} + y^{2} + z^{2} = \sqrt{2} z (x + y) \end{cases}

Answered by MJS_new last updated on 21/Jun/22

let y = p x \land z = q x

{\begin{cases} (3 p^{2} + q^{2} + 2) x^{2} - 7 = 0 \\ (p^{2} - \sqrt{2} p q + q^{2} - \sqrt{2} q + 1) x^{2} = 0 \end{cases}

\Rightarrow x = 0 \lor (p^{2} - \sqrt{2} p q + q^{2} - \sqrt{2} q + 1) = 0

if x = 0 the first equation is wrong and

p^{2} - \sqrt{2} p q + q^{2} - \sqrt{2} q + 1 = 0 \Rightarrow q = \frac{\sqrt{2}}{2} (p + 1 \pm (p - 1) i)

\Rightarrow only real if p = 1 \Rightarrow q = \sqrt{2}

\Rightarrow (3 + 2 + 2) x^{2} - 7 = 0 \Rightarrow x = \pm 1

\Rightarrow x = \pm 1 \land y = x \land z = \sqrt{2} x

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