Solve-for-real-numbers-x-y-5-3-x-3-1-5-y-5-6-1-5-1-

Question Number 158187 by HongKing last updated on 31/Oct/21

Solve for real numbers :

{\begin{cases} \sqrt{x} - y^{5} = 3 \\ \sqrt[5]{\sqrt{x} - 3} - \sqrt[5]{y^{5} + 6} = - 1 \end{cases}

Answered by MJS_new last updated on 31/Oct/21

\sqrt{x} = y^{5} + 3 [x ⩾ 0 \land \sqrt{x} ⩾ 0]

y - {(y^{5} + 6)}^{1 / 5} = - 1

y^{4} + 2 y^{3} + 2 y^{2} + y - 1 = 0

(y^{2} + y + \frac{1 - \sqrt{5}}{2}) (y^{2} + y + \frac{1 + \sqrt{5}}{2}) = 0

y = \frac{- 1 + \sqrt{- 1 + 2 \sqrt{5}}}{2}

\Rightarrow x = \frac{7 + 5 \sqrt{5}}{2}

Commented by HongKing last updated on 31/Oct/21

perfect dear S er, thank you so much

Commented by ajfour last updated on 01/Nov/21

a w e s t r i k e p r e c i s i o n!

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