If A.M and G.M of x and y are in the ratio p | vedclass

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Question

(b) $\frac{{\frac{{x + y}}{2}}}{{\sqrt {xy} }} = \frac{p}{q}$

$\frac{{x + y}}{{2(\sqrt {xy} )}} = \frac{p}{q}$&hellip;..$(i)$

$\frac{{{x^2} + {y

Vedclass · Accepted Answer

$p + \sqrt {{p^2} - {q^2}} $:$p - \sqrt {{p^2} - {q^2}} $

Vedclass · Answer

(b) $\frac{{\frac{{x + y}}{2}}}{{\sqrt {xy} }} = \frac{p}{q}$

$\frac{{x + y}}{{2(\sqrt {xy} )}} = \frac{p}{q}$&hellip;..$(i)$

$\frac{{{x^2} + {y^2} + 2xy}}{{4xy}} = \frac{{{p^2}}}{{{q^2}}}$

$\frac{{{x^2} + {y^2} + 2xy - 4xy}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{{{(x - y)}^2}}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{x - y}}{{2\sqrt {xy} }} = \frac{{\sqrt {{p^2} - {q^2}} }}{q}$&hellip;..$(ii)$

Equation $(i)$ is divided by $(ii),$

Then $\frac{{x + y}}{{x - y}} = \frac{p}{{\sqrt {{p^2} - {q^2}} }}$;

$\frac{x}{y} = \frac{{p + \sqrt {{p^2} - {q^2}} }}{{p - \sqrt {{p^2} - {q^2}} }}$.

If $A.M$ and $G.M$ of $x$ and $y$ are in the ratio $p : q$, then $x : y$ is

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