If $(\sqrt{a} + \sqrt{b})$ $\left( {\sqrt a + \sqrt b } \right)$ = 15 and $(\sqrt{a} - \sqrt{b})$ $\left( {\sqrt a - \sqrt b } \right)$ = 3, then the value of $\frac{\sqrt{a b}}{4}$ $\frac{{\sqrt {ab} }}{4}$ is?

16
27
3 $\frac{27}{2}$ $\frac{{27}}{2}$
45

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (\sqrt{a} + \sqrt{b}) = 15 \\ Square both sides \\ \Rightarrow a + b + 2 \sqrt{a b} = 225 \\ \Rightarrow a + b = 225 - 2 \sqrt{a b} . . . . . (i) \\ (\sqrt{a} - \sqrt{b}) = 3 \\ Square both sides \\ \Rightarrow a + b - 2 \sqrt{a b} = 9 \\ \Rightarrow a + b = 9 + 2 \sqrt{a b} . . . . . (i i) \\ From (i) and (ii) \\ \Rightarrow 225 - 2 \sqrt{a b} = 9 + 2 \sqrt{a b} \\ \Rightarrow 216 = 4 \sqrt{a b} \\ \Rightarrow 54 = \sqrt{a b} \\ Divided by 4 on both sides \\ \Rightarrow \frac{\sqrt{a b}}{4} = \frac{54}{4} \\ \Rightarrow \frac{\sqrt{a b}}{4} = \frac{27}{2} \end{aligned}

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