x-4-ax-2-by-2-y-4-bx-2-ay-2-solve-for-x-y-a-b-R-a-b-0-

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!