If

(a−−√+b√)$$\left(\sqrt{a}+\sqrt{b}\right)$$

$$\left( {\sqrt a + \sqrt b } \right)$$   = 15 and

(a−−√−b√)$$\left(\sqrt{a}-\sqrt{b}\right)$$

$$\left( {\sqrt a - \sqrt b } \right)$$   = 3, then the value of

ab−−√4$$\frac{\sqrt{ab}}{4}$$

$$\frac{{\sqrt {ab} }}{4}$$   is?

Explanation:-

$$\eqalign{ & \left( {\sqrt a + \sqrt b } \right) = 15{\text{ }} \cr & {\text{Square both sides}} \cr & \Rightarrow a + b + 2\sqrt {ab} = 225 \cr & \Rightarrow a + b = 225 - 2\sqrt {ab} \,.....(i) \cr & {\text{ }}\left( {\sqrt a - \sqrt b } \right) = 3 \cr & {\text{Square both sides}} \cr & \Rightarrow a + b - 2\sqrt {ab} = 9 \cr & \Rightarrow a + b = 9 + 2\sqrt {ab} \,.....(ii) \cr & {\text{From (i) and (ii)}} \cr & \Rightarrow 225 - 2\sqrt {ab} = 9 + 2\sqrt {ab} \cr & \Rightarrow 216 = 4\sqrt {ab} \cr & \Rightarrow 54 = \sqrt {ab} \cr & {\text{Divided by 4 on both sides}} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{54}}{4} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{27}}{2} \cr}$$