Solve-for-real-numbers-x-32-x-16-y-2-2-2-x-12-y-

Question Number 158275 by HongKing last updated on 01/Nov/21

Solve for real numbers :

x^{32} + x^{16} + y^{2} = 2 \sqrt{2} x^{12} y

Answered by mindispower last updated on 01/Nov/21

x^{32} + x^{16} ⩾ 2 x^{24} e q u a l i t e x^{16} = x^{8} \Rightarrow x \in {0, 1, - 1

2 x^{24} + y^{2} ⩾ 2 \sqrt{2} x^{12} y

\Leftrightarrow {(\sqrt{2} x^{12} - y)}^{2} ⩾ 0

y = \sqrt{2} x^{12} \in {0, \sqrt{2}}

(x, y) = {(0, 0); (0, \sqrt{2}); (1, 0); (1, \sqrt{2}); (- 1, 0); (- 1, \sqrt{2})}

Commented by HongKing last updated on 02/Nov/21

cool my dear Ser, thank you

Answered by MJS_new last updated on 02/Nov/21

we can easily solve for y :

y = x^{8} (\sqrt{2} x^{4} + ∣ x^{8} - 1 ∣ i)

\Rightarrow to stay in R we need x^{8} = 0 \lor x^{8} - 1 = 0 \Rightarrow

\Rightarrow x = 0 \lor x = \pm 1

\Rightarrow

x = 0 \land y = 0 \lor x = \pm 1 \land y = \sqrt{2}

Commented by HongKing last updated on 02/Nov/21

cool my dear S ir, thank you so much