Question

Solve the following equations:
$6x^4+x^2{y}^2+16=2x(12x+y^3),x^2+xy-y^2=4$.

Answer 1

$⟹$ From $x^2+xy-y^2-4=0$

$x=\frac{-y\pm \sqrt{y^2+4(y^2+4)}}{2}$
$=\frac{-y\pm \sqrt{{5y}^2+16}}{2}$
Putting
$x=\frac{-y+\sqrt{5y^2+16}}{2}$ in equation $\left(I\right)$
$6{\left(\frac{-y+\sqrt{5y^2+16}}{2}\right)}^4+{\left(\frac{-y+\sqrt{5y^2+16}}{2}\right)}^2+$

$16-24{\left(\frac{-y+\sqrt{5y^2+16}}{2}\right)}^2-2\cdot \left(\frac{-y+\sqrt{5y^2+16}}{2}\right)y^3=0$
On solving above equation, we get $y=2$
Similarly putting:-
$x=\frac{-y-\sqrt{5y^2+16}}{2}$ in equation $\left(I\right)$ we get $y=-2$
Now, substituting $y=2$ in equation $\left(II\right):-$
$x^2+2x-8=0$
$(x+4)(x-2)=0$
$x=2,-4$
Substituting $y=-2$ in equation $\left(II\right):-$
$x^2+2x-8=0$
$(x-4)(x+2)=0$
$x=-2,4$
$\therefore$Solutions are
$x=-2,y=-2$
$x=4,y=-2$
$x=2,y=2$
$x=-4,y=2$