Solve the following system of equations: $\;$ $\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$, $\;$ $x^2 + y^2 = 15$
Given system of equations:
$\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^2 + y^2 = 15$ $\;\;\; \cdots \; (2)$
Let $\;$ $\sqrt{\dfrac{x}{y}} = p$ $\;\;\; \cdots \; (3a)$
Then $\;$ $\sqrt{\dfrac{y}{x}} = \dfrac{1}{p}$ $\;\;\; \cdots \; (3b)$
In view of equations $(3a)$ and $(3b)$, equation $(1)$ becomes
$p + \dfrac{1}{p} = \dfrac{5}{2}$
i.e. $\;$ $p^2 + 1 = \dfrac{5}{2} p$
i.e. $\;$ $2p^2 - 5p + 2 = 0$
i.e. $\;$ $\left(p - 2\right) \left(2p - 1\right) = 0$
i.e. $\;$ $p = 2$ $\;\;$ or $\;\;$ $p = \dfrac{1}{2}$
Substituting the values of $p$ in equation $(3a)$ give
Case 1:
When $\;$ $p = 2$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = 2$
i.e. $\;$ $\dfrac{x}{y} = 4$
i.e. $\;$ $x = 4y$ $\;\;\; \cdots \; (4a)$
Substituting $\;$ $x = 4y$ $\;$ in equation $(2)$ gives
$\left(4y\right)^2 + y^2 = 15$
i.e. $\;$ $17 y^2 = 15$ $\implies$ $y = \pm \sqrt{\dfrac{15}{17}}$
Substituting the value of $y$ in equation $(4a)$ gives $\;\;\;$ $x = \pm 4 \sqrt{\dfrac{15}{17}}$
Case 2:
When $\;$ $p = \dfrac{1}{2}$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{x}{y} = \dfrac{1}{4}$
i.e. $\;$ $y = 4x$ $\;\;\; \cdots \; (4b)$
Substituting $\;$ $y = 4x$ $\;$ in equation $(2)$ gives
$x^2 + \left(4x\right)^2 = 15$
i.e. $\;$ $17 x^2 = 15$ $\implies$ $x = \pm \sqrt{\dfrac{15}{17}}$
Substituting the value of $x$ in equation $(4b)$ gives $\;\;\;$ $y = \pm 4 \sqrt{\dfrac{15}{17}}$
$\therefore \;$ The solution to the given system of equations is
$\left(x, y\right) =$
$ \left\{\left(\dfrac{4 \sqrt{15}}{\sqrt{17}}, \dfrac{\sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{-4 \sqrt{15}}{\sqrt{17}}, \dfrac{-\sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{\sqrt{15}}{\sqrt{17}}, \dfrac{4 \sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{- \sqrt{15}}{\sqrt{17}}, \dfrac{-4 \sqrt{15}}{\sqrt{17}}\right) \right\}$