x^4 =ax^2 +by^2 y^4 =bx^2 +ay^2 solve for x, y. [a ,b∈ R; a, b≠0]


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Question Number 50825 by behi83417@gmail.com last updated on 20/Dec/18

$x^{4} = {ax}^{2} + {by}^{2}$ $y^{4} = {bx}^{2} + {ay}^{2}$ $solve for x, y . [a, b \in R; a, b \neq 0]$
	Answered by mr W last updated on 21/Dec/18

	$x = y = 0 i s a s o l u t i o n .$ $l e t X = x^{2}, Y = y^{2}$ $X^{2} = a X + b Y$ $Y^{2} = b X + a Y$ ${(X + Y)}^{2} - 2 X Y = (a + b) (X + Y)$ ${(X Y)}^{2} = a b (X^{2} + Y^{2}) + (a^{2} + b^{2}) X Y$ ${(X Y)}^{2} = a b (X^{2} + Y^{2} + 2 X Y) + (a^{2} + b^{2} - 2 a b) X Y$ ${(X Y)}^{2} = a b {(X + Y)}^{2} + {(a - b)}^{2} X Y$ $l e t u = X + Y, v = X Y$ $u^{2} - (a + b) u - 2 v = 0$ $\Rightarrow v = \frac{[u - (a + b)] u}{2}$ ${a b u}^{2} - v^{2} + {(a - b)}^{2} v = 0$ ${a b u}^{2} - {[\frac{u^{2} - (a + b) u}{2}]}^{2} + {(a - b)}^{2} \frac{u^{2} - (a + b) u}{2} = 0$ $4 {a b u}^{2} - {[u^{2} - (a + b) u]}^{2} + 2 {(a - b)}^{2} [u^{2} - (a + b) u] = 0$ $u \neq 0 (u = 0 \Rightarrow x = y = 0)$ $4 a b u - u^{2} + 2 (a + b) u - {(a + b)}^{2} + 2 {(a - b)}^{2} u - 2 {(a - b)}^{2} (a + b) = 0$ $u^{2} - 2 (a + b + a^{2} + b^{2}) u + [a + b + 2 {(a - b)}^{2}] (a + b) = 0$ $\Rightarrow u = (a + b + a^{2} + b^{2}) \pm \sqrt{{(a + b + a^{2} + b^{2})}^{2} - (a + b) [a + b + 2 {(a - b)}^{2}]}$ $\Rightarrow v = \frac{[u - (a + b)] u}{2}$ $X + Y = u$ $X Y = v$ $\Rightarrow X a n d Y a r e r o o t s o f$ $t^{2} - u t + v = 0$ $\Rightarrow X, Y = \frac{u \pm \sqrt{u^{2} - 4 v}}{2}$ $\Rightarrow x, y = \pm \sqrt{\frac{u \pm \sqrt{u^{2} - 4 v}}{2}}$
	Commented by behi83417@gmail.com last updated on 21/Dec/18

	$t h a n k y o u s o m u c h d e a r m a s t e r .$ $p e r f e c t!$
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