Solve :

$\sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2}$

Let $\sqrt{\dfrac{x}{x - 3}} =$ a …….(i)

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 = 5a \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.$

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = \dfrac{1}{2}$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = \dfrac{1}{4} \\[1em] \Rightarrow 4x = x - 3 \\[1em] \Rightarrow 4x - x = -3 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1.$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = 2$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = 4 \\[1em] \Rightarrow x = 4(x - 3) \\[1em] \Rightarrow x = 4x - 12 \\[1em] \Rightarrow 4x - x = 12 \\[1em] \Rightarrow 3x = 12 \\[1em] \Rightarrow x = 4.$

Hence, x = -1, 4.