Solve the following system of equations: $\;$ $x^2 + y^2 = 20, \;\; xy = 8$
Given system of equations:
$x^2 + y^2 = 20$ $\;\;\; \cdots \; (1)$
and $\;$ $xy = 8$ $\;$ i.e. $\;$ $x = \dfrac{8}{y}$ $\;\;\; \cdots \; (2)$
Then, in view of equation $(2)$ equation $(1)$ becomes
$\left(\dfrac{8}{y}\right)^2 + y^2 = 20$
i.e. $\;$ $64 + y^4 = 20 y^2$
i.e. $\;$ $y^4 - 20y^2 + 64 = 0$
i.e. $\;$ $y^4 - 16y^2 - 4y^2 + 64 = 0$
i.e. $\;$ $y^2 \left(y^2 - 4\right) - 16 \left(y^2 - 4\right) = 0$
i.e. $\;$ $\left(y^2 - 16\right) \left(y^2 - 4\right) = 0$
i.e. $\;$ $y^2 = 16$ $\;$ or $\;$ $y^2 = 4$
i.e. $\;$ $y = \pm 4$ $\;$ or $\;$ $y = \pm 2$
When $\;$ $y = \pm 4$, $\;$ we have from equation $(2)$,
$x = \dfrac{8}{\pm 4} = \pm 2$
When $\;$ $y = \pm 2$, $\;$ we have from equation $(2)$,
$x = \dfrac{8}{\pm 2} = \pm 4$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(2, 4\right), \; \left(-2, -4\right), \; \left(4, 2\right), \; \left(-4, -2\right) \right\}$