Question

The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{be}\mathit{}a\mathrm{and}\mathit{}b. \\ \mathrm{Let}\mathrm{the}\mathrm{geometric}\mathrm{mean}\mathrm{between}\mathrm{them}\mathrm{be}G. \\ \mathrm{We}\mathrm{have}: \\ a+b=6G \\ \mathrm{But},G=\sqrt{ab} \\ \therefore a+b=6\sqrt{ab} \\ \Rightarrow {\left(a+b\right)}^2={\left(6\sqrt{ab}\right)}^2 \\ \Rightarrow a^2+2ab+b^2=36ab \\ \Rightarrow a^2-34ab+b^2=0 \\ \mathrm{Using}\mathrm{the}\mathrm{quadratic}\mathrm{formula}: \\ \Rightarrow a=\frac{-\left(-34b\right)\pm \sqrt{{\left(-34b\right)}^2-4\times 1\times b^2}}{2\times 1} \\ \Rightarrow a=\frac{34b\pm b\sqrt{1156-4}}{2} \\ \Rightarrow a=\frac{b\left(34\pm \sqrt{1152}\right)}{2} \\ \Rightarrow \frac{a}{b}=\frac{34\pm 24\sqrt{2}}{2} \\ \Rightarrow \frac{a}{b}=17+12\sqrt{2}\left[\because a\mathrm{and}b\mathrm{are}\mathrm{positive}\mathrm{numbers}\right] \\ \Rightarrow \frac{a}{b}=3+8+2\times 3\times 2\sqrt{2} \\ \Rightarrow \frac{a}{b}={\left(3+2\sqrt{2}\right)}^2 \\ \Rightarrow \frac{a}{b}=\frac{{\left(3+2\sqrt{2}\right)}^2\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)} \\ \Rightarrow \frac{a}{b}=\frac{\left(3+2\sqrt{2}\right)\left(9-8\right)}{\left(3-2\sqrt{2}\right)} \\ \Rightarrow \frac{a}{b}=\frac{\left(3+2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)} \\ \Rightarrow a:b=\left(3+2\sqrt{2}\right):\left(3-2\sqrt{2}\right) \end{gathered}$$