Question

If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that:

\[a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}) .\]

Answer 1

AM = 2GM

\[ \therefore \frac{a + b}{2} = 2\sqrt{ab}\]

\[ \Rightarrow a + b = 4\sqrt{ab}\]

Squaring both the sides : 

\[ \Rightarrow \left( a + b \right)^2 = \left( 4\sqrt{ab} \right)^2 \]

\[ \Rightarrow a^2 + 2ab + b^2 = 16ab\]

\[ \Rightarrow a^2 - 14ab + b^2 = 0\]

Using the quadratic formula: 

\[ \Rightarrow a = \frac{- \left( - 14b \right) + \sqrt{\left( - 14b \right)^2 - 4 \times 1 \times b^2}}{2 \times 1} \left[ \because\text {  a is positive number } \right]\]

\[ \Rightarrow a = \frac{14b + 2b\sqrt{49 - 1}}{2}\]

\[ \Rightarrow a = b\left( 7 + 4\sqrt{3} \right)\]

\[ \Rightarrow \frac{a}{b} = 7 + 4\sqrt{3}\]

\[ \Rightarrow \frac{a}{b} = 4 + 3 + 2 \times 2 \times \sqrt{3}\]

\[ \Rightarrow \frac{a}{b} = \left( 2 + \sqrt{3} \right)^2 \]

\[ \Rightarrow \frac{a}{b} = \frac{\left( 2 + \sqrt{3} \right)^2 \left( 2 - \sqrt{3} \right)}{\left( 2 - \sqrt{3} \right)}\]

\[ \Rightarrow \frac{a}{b} = \frac{\left( 2 + \sqrt{3} \right)\left( 4 - 3 \right)}{\left( 2 - \sqrt{3} \right)}\]

\[ \therefore \frac{a}{b} = \frac{\left( 2 + \sqrt{3} \right)}{\left( 2 - \sqrt{3} \right)}\]