Question : If $a+b :\sqrt{ab} = 4:1 $ where $ a > b > 0$, then $ a:b$ is:
Option 1: $\left (2+\sqrt{3} \right):\left (2-\sqrt{3} \right)$
Option 2: $\left (2-\sqrt{3} \right):\left (2+\sqrt{3} \right)$
Option 3: $\left (3+\sqrt{2} \right):\left (3-\sqrt{2} \right)$
Option 4: $\left (3-\sqrt{2} \right):\left (3+\sqrt{2} \right)$
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Correct Answer: $\left (2+\sqrt{3} \right):\left (2-\sqrt{3} \right)$
Solution :
Given:
$\frac{\left (a+b \right)}{\sqrt{ab}}=\frac{4}{1}$
⇒ $\frac{a+b}{2\sqrt{ab}}=\frac{2}{1}$
Applying componendo and dividendo, we get,
⇒ $\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}$
⇒ $\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{3}{1}$
⇒ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$
Again applying componendo and dividendo we get,
⇒ $\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
⇒ $\frac{a}{b}=\left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )^2$
⇒ $\frac{a}{b}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}$
⇒ $\frac{a}{b}=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}$
$\therefore \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Hence, the correct answer is $({2+\sqrt{3}}):({2-\sqrt{3}})$.
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