A simple system of equations

Given that:

$\begin{cases} a + b = \dfrac {25}4 \\ (1 + \sqrt a)(1+\sqrt b) = \dfrac {15}2 \end{cases} \implies \begin{cases} (\sqrt a + \sqrt b)^2 - 2\sqrt{ab} = \dfrac {25}4 \\ 1 + \sqrt a + \sqrt b + \sqrt{ab} = \dfrac {15}2 \end{cases} \\ \implies \begin{cases} (\sqrt a + \sqrt b)^2 = \dfrac {25}4 + 2\sqrt{ab} & ...(1) \\ \sqrt a + \sqrt b = \dfrac {13}2 - \sqrt{ab} & ...(2) \end{cases}$

From $(2)^2 = (1)$ :

$\begin{aligned} \left(\frac {13}2 - \sqrt{ab}\right)^2 & = \frac {25}4 + 2\sqrt{ab} \\ ab - 13\sqrt{ab} + \frac {169}4 & = \frac {25}4 + 2\sqrt{ab} \\ ab - 15 \sqrt{ab} + 36 & = 0 & \small \blue{\text{A quadratic equation of }\sqrt{ab}} \\ (\sqrt{ab}-3)(\sqrt{ab}-12) & = 0 \end{aligned}$

$\implies \begin{cases} \sqrt{ab} = 3 & \implies ab = \blue{\boxed 9} \\ \sqrt{ab} = 12 & \implies (2): \ \sqrt a + \sqrt b = \dfrac {13}2 - 12 < 0, \red{\text{ No real solution}} \end{cases}$