Question

Solve the equation.
$x^2-xy+y^2=7$, $x^4+x^2{y}^2+y^4=133$.

Answer 1

From the equation, we have
$(x^2+y^2{)}^2-x^2{y}^2=133$
or $(x^2+y^2-xy)(x^2+y^2+xy)=133$
$\therefore x^2+y^2+xy=133/7=19$
$x^2+y^2-xy=7$(given)
Adding and subtracting, we get
$x^2+y^2=13,xy=16$
$\therefore x^2{y}^2=36$
$\therefore x^2,y^2$ are the roots of $t^2-13t+36=0$
$\therefore (t-4)(t-9)=0$ $\therefore t=4,9$
$x^2=4,y^2=9$ or $x^2=9,y^2=4$
$\therefore x=3,y=2$ or $x=-3,y=-2$
Similarly $x=2$, $y=3$ or $x=-2$, $y=-3$
Note: We have taken the roots keeping in view that $xy=6$.