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

$$\begin{gathered} \mathrm{AM}=2\mathrm{GM} \\ \therefore \frac{a+b}{2}=2\sqrt{ab} \\ \Rightarrow a+b=4\sqrt{ab} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \Rightarrow {\left(a+b\right)}^2={\left(4\sqrt{ab}\right)}^2 \\ \Rightarrow a^2+2ab+b^2=16ab \\ \Rightarrow a^2-14ab+b^2=0 \\ \mathrm{Using}\mathrm{the}\mathrm{quadratic}\mathrm{formula}: \\ \Rightarrow a=\frac{-\left(-14b\right)+\sqrt{{\left(-14b\right)}^2-4\times 1\times b^2}}{2\times 1}\left[\because a\mathrm{is}\mathrm{positive}\mathrm{number}\right] \\ \Rightarrow a=\frac{14b+2b\sqrt{49-1}}{2} \\ \Rightarrow a=b\left(7+4\sqrt{3}\right) \\ \Rightarrow \frac{a}{b}=7+4\sqrt{3} \\ \Rightarrow \frac{a}{b}=4+3+2\times 2\times \sqrt{3} \\ \Rightarrow \frac{a}{b}={\left(2+\sqrt{3}\right)}^2 \\ \Rightarrow \frac{a}{b}=\frac{{\left(2+\sqrt{3}\right)}^2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)} \\ \Rightarrow \frac{a}{b}=\frac{\left(2+\sqrt{3}\right)\left(4-3\right)}{\left(2-\sqrt{3}\right)} \\ \therefore \frac{a}{b}=\frac{\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)} \end{gathered}$$