Question

The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.

Answer 1

Let the two numbers be a and b. 

Let the geometric mean between them be G. 

We have: 

a + b = 6G

\[\text { But }, G = \sqrt{ab}\]

\[ \therefore a + b = 6\sqrt{ab}\]

\[ \Rightarrow \left( a + b \right)^2 = \left( 6\sqrt{ab} \right)^2 \]

\[ \Rightarrow a^2 + 2ab + b^2 = 36ab\]

\[ \Rightarrow a^2 - 34ab + b^2 = 0\]

\[\text { Using the quadratic formula: } \]

\[ \Rightarrow a = \frac{- \left( - 34b \right) \pm \sqrt{\left( - 34b \right)^2 - 4 \times 1 \times b^2}}{2 \times 1}\]

\[ \Rightarrow a = \frac{34b \pm b\sqrt{1156 - 4}}{2}\]

\[ \Rightarrow a = \frac{b\left( 34 \pm \sqrt{1152} \right)}{2}\]

\[ \Rightarrow \frac{a}{b} = \frac{34 \pm 24\sqrt{2}}{2}\]

\[ \Rightarrow \frac{a}{b} = 17 + 12\sqrt{2} \left[ \because \text { a and b are positive numbers } \right]\]

\[ \Rightarrow \frac{a}{b} = 3 + 8 + 2 \times 3 \times 2\sqrt{2}\]

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

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

\[ \Rightarrow \frac{a}{b} = \frac{\left( 3 + 2\sqrt{2} \right)\left( 9 - 8 \right)}{\left( 3 - 2\sqrt{2} \right)}\]

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

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