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∈ R; a, b≠0]](https://www.tinkutara.com/question/Q50825.png)
$$\boldsymbol{\mathrm{x}}^{\mathrm{4}} =\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\boldsymbol{\mathrm{by}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{y}}^{\mathrm{4}} =\boldsymbol{\mathrm{bx}}^{\mathrm{2}} +\boldsymbol{\mathrm{ay}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\:\boldsymbol{\mathrm{y}}.\:\left[\boldsymbol{\mathrm{a}}\:,\boldsymbol{\mathrm{b}}\in\:\boldsymbol{\mathrm{R}};\:\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}}\neq\mathrm{0}\right] \\ $$
Answered by mr W last updated on 21/Dec/18
![x=y=0 is a solution. let X=x^2 , Y=y^2 X^2 =aX+bY Y^2 =bX+aY (X+Y)^2 −2XY=(a+b)(X+Y) (XY)^2 =ab(X^2 +Y^2 )+(a^2 +b^2 )XY (XY)^2 =ab(X^2 +Y^2 +2XY)+(a^2 +b^2 −2ab)XY (XY)^2 =ab(X+Y)^2 +(a−b)^2 XY letu=X+Y, v=XY u^2 −(a+b)u−2v=0 ⇒v=(([u−(a+b)]u)/2) abu^2 −v^2 +(a−b)^2 v=0 abu^2 −[((u^2 −(a+b)u)/2)]^2 +(a−b)^2 ((u^2 −(a+b)u)/2)=0 4abu^2 −[u^2 −(a+b)u]^2 +2(a−b)^2 [u^2 −(a+b)u]=0 u≠0 (u=0 ⇒x=y=0) 4abu−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 ⇒u=(a+b+a^2 +b^2 )±(√((a+b+a^2 +b^2 )^2 −(a+b)[a+b+2(a−b)^2 ])) ⇒v=(([u−(a+b)]u)/2) X+Y=u XY=v ⇒X and Y are roots of t^2 −ut+v=0 ⇒X,Y=((u±(√(u^2 −4v)))/2) ⇒x,y=±(√((u±(√(u^2 −4v)))/2))](https://www.tinkutara.com/question/Q50828.png)
$${x}={y}=\mathrm{0}\:{is}\:{a}\:{solution}. \\ $$$$ \\ $$$${let}\:{X}={x}^{\mathrm{2}} ,\:{Y}={y}^{\mathrm{2}} \\ $$$${X}^{\mathrm{2}} ={aX}+{bY} \\ $$$${Y}^{\mathrm{2}} ={bX}+{aY} \\ $$$$\left({X}+{Y}\right)^{\mathrm{2}} −\mathrm{2}{XY}=\left({a}+{b}\right)\left({X}+{Y}\right) \\ $$$$\left({XY}\right)^{\mathrm{2}} ={ab}\left({X}^{\mathrm{2}} +{Y}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){XY} \\ $$$$\left({XY}\right)^{\mathrm{2}} ={ab}\left({X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +\mathrm{2}{XY}\right)+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right){XY} \\ $$$$\left({XY}\right)^{\mathrm{2}} ={ab}\left({X}+{Y}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} {XY} \\ $$$${letu}={X}+{Y},\:{v}={XY} \\ $$$${u}^{\mathrm{2}} −\left({a}+{b}\right){u}−\mathrm{2}{v}=\mathrm{0} \\ $$$$\Rightarrow{v}=\frac{\left[{u}−\left({a}+{b}\right)\right]{u}}{\mathrm{2}} \\ $$$${abu}^{\mathrm{2}} −{v}^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} {v}=\mathrm{0} \\ $$$${abu}^{\mathrm{2}} −\left[\frac{{u}^{\mathrm{2}} −\left({a}+{b}\right){u}}{\mathrm{2}}\right]^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} \frac{{u}^{\mathrm{2}} −\left({a}+{b}\right){u}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{4}{abu}^{\mathrm{2}} −\left[{u}^{\mathrm{2}} −\left({a}+{b}\right){u}\right]^{\mathrm{2}} +\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} \left[{u}^{\mathrm{2}} −\left({a}+{b}\right){u}\right]=\mathrm{0} \\ $$$${u}\neq\mathrm{0}\:\left({u}=\mathrm{0}\:\Rightarrow{x}={y}=\mathrm{0}\right) \\ $$$$\mathrm{4}{abu}−{u}^{\mathrm{2}} +\mathrm{2}\left({a}+{b}\right){u}−\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} {u}−\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)=\mathrm{0} \\ $$$${u}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){u}+\left[{a}+{b}+\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} \right]\left({a}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\left({a}+{b}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\pm\sqrt{\left({a}+{b}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({a}+{b}\right)\left[{a}+{b}+\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} \right]} \\ $$$$\Rightarrow{v}=\frac{\left[{u}−\left({a}+{b}\right)\right]{u}}{\mathrm{2}} \\ $$$${X}+{Y}={u} \\ $$$${XY}={v} \\ $$$$\Rightarrow{X}\:{and}\:{Y}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} −{ut}+{v}=\mathrm{0} \\ $$$$\Rightarrow{X},{Y}=\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}}}{\mathrm{2}} \\ $$$$\Rightarrow{x},{y}=\pm\sqrt{\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}}}{\mathrm{2}}} \\ $$
Commented by behi83417@gmail.com last updated on 21/Dec/18
$${thank}\:{you}\:{so}\:{much}\:{dear}\:{master}. \\ $$$${perfect}! \\ $$