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

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Question

(b) Given $A.M.$$ = 2$($G.M.$)

or $\frac{1}{2}(a + b) = 2\sqrt {ab} $

or $\frac{{a + b}}{{2\sqrt {ab} }} = \frac{2}{1}$

$ \Rightarrow $$\frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Given $A.M.$$ = 2$($G.M.$)

or $\frac{1}{2}(a + b) = 2\sqrt {ab} $

or $\frac{{a + b}}{{2\sqrt {ab} }} = \frac{2}{1}$

$ \Rightarrow $$\frac{{a + b + 2\sqrt {ab} }}{{a + b - 2\sqrt {ab} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{{{(\sqrt a + \sqrt b )}^2}}}{{{{(\sqrt a - \sqrt b )}^2}}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} = \frac{{\sqrt 3 }}{1}$

$ \Rightarrow $ $\frac{a}{b} = {\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}$

$ \Rightarrow $ $\frac{a}{b} = \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}$

or $a:b = (2 + \sqrt 3 ):(2 - \sqrt 3 )$.

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

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