If the A.M. is twice the G.M. of the numbers | vedclass

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(B) Given $A.M. = 2(G.M.)$.$\frac{a + b}{2} = 2\sqrt{ab}$.$\frac{a + b}{\sqrt{ab}} = 4$.Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$ and $\f

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$\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$

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(B) Given $A.M. = 2(G.M.)$.$\frac{a + b}{2} = 2\sqrt{ab}$.$\frac{a + b}{\sqrt{ab}} = 4$.Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$ and $\frac{b}{a} = \frac{1}{x^2}$.Dividing by $\sqrt{ab}$,we get $\sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = 4$.$x + \frac{1}{x} = 4$.$x^2 - 4x + 1 = 0$.Using the quadratic formula,$x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.Since $\frac{a}{b} = x^2$,we have $\frac{a}{b} = (2 \pm \sqrt{3})^2 = 4 + 3 \pm 4\sqrt{3} = 7 \pm 4\sqrt{3}$.Alternatively,using the componendo and dividendo rule on $\frac{a+b}{2\sqrt{ab}} = \frac{2}{1}$:$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{2+1}{2-1} = 3$.$\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 3$.$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{3}$.Applying componendo and dividendo again:$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.$\sqrt{\frac{a}{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.$\frac{a}{b} = \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}} = \frac{4+2\sqrt{3}}{4-2\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}}$.

If the $A.M.$ is twice the $G.M.$ of the numbers $a$ and $b$,then $a:b$ will be

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