Solve the following system of equations: $\;$ $\sqrt{\dfrac{x}{y}} - \sqrt{\dfrac{y}{x}} = \dfrac{3}{2}$, $\;$ $x + y + xy = 9$
Given system of equations:
$\sqrt{\dfrac{x}{y}} - \sqrt{\dfrac{y}{x}} = \dfrac{3}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y + xy = 9$ $\;\;\; \cdots \; (2)$
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
$p - \dfrac{1}{p} = \dfrac{3}{2}$
i.e. $\;$ $p^2 - 1 = \dfrac{3}{2} p$
i.e. $\;$ $2p^2 - 3p - 2 = 0$
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
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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
$4y + y + 4y^2 = 9$
i.e. $\;$ $4y^2 + 5y - 9 = 0$
i.e. $\;$ $\left(y - 1\right) \left(4y + 9\right) = 0$
i.e. $\;$ $y = 1$ $\;\;$ or $\;\;$ $y = \dfrac{-9}{4}$
Substituting the value of $y$ in equation $(4a)$ gives
- when $\;$ $y = 1$, $\;$ $x = 4 \times 1 = 4$
- when $\;$ $y = \dfrac{-9}{4}$, $\;$ $x = 4 \times \left(\dfrac{-9}{4}\right) = -9$
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when $\;$ $p = \dfrac{-1}{2}$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = \dfrac{-1}{2}$
But square root of any number cannot be negative.
$\implies$ $p = \dfrac{-1}{2}$ $\;$ is not an acceptable solution.