Question

The Sum of two no. is 6 times their geometric mean, show that no. are in the ratio (3 + 3 √2) : (3 – 2 √2)​

Answer 1

Appropriate Question :-

The sum of two positive numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) : (3 – 2 √2)

$\large\underline{\sf{Solution-}}$

Let assume that

• Two positive numbers be x and y.

So, According to statement,

The Sum of two no. is 6 times their geometric mean.

$\rm \implies\:x + y = 6 \sqrt{xy}$

can be further rewritten as

$\rm :\longmapsto\:x + y = 3 \times 2 \sqrt{xy}$

$\rm :\longmapsto\:\dfrac{x + y}{2 \sqrt{xy} } =3$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x + y}{2 \sqrt{xy} } =\dfrac{3}{1}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{x + y + 2 \sqrt{xy} }{x + y - 2 \sqrt{xy} } =\dfrac{3 + 1}{3 - 1}$

$\rm :\longmapsto\:\dfrac{ {( \sqrt{x} )}^{2} + {( \sqrt{y} )}^{2} + 2 \sqrt{xy} }{ {( \sqrt{x} )}^{2} + {( \sqrt{y}) }^{2} - 2 \sqrt{xy} } =\dfrac{4}{2}$

$\rm :\longmapsto\:\dfrac{ {( \sqrt{x} + \sqrt{y} )}^{2} }{ {( \sqrt{x} - \sqrt{y})}^{2} } = 2$

$\rm \implies\:\dfrac{ \sqrt{x} + \sqrt{y} }{ \sqrt{x} - \sqrt{y} } = \sqrt{2}$

can further rewritten as

$\rm :\longmapsto\:\dfrac{ \sqrt{x} + \sqrt{y} }{ \sqrt{x} - \sqrt{y} } = \dfrac{ \sqrt{2} }{1}$

On applying for and Dividendo, we get

$\rm :\longmapsto\:\dfrac{ \sqrt{x} + \sqrt{y} + \sqrt{x} - \sqrt{y} }{ \sqrt{x} - \sqrt{y} - \sqrt{x} - \sqrt{y} } = \dfrac{ \sqrt{2} + 1}{ \sqrt{2} - 1}$

$\rm :\longmapsto\:\dfrac{ 2\sqrt{x}}{ 2\sqrt{y} } = \dfrac{ \sqrt{2} + 1}{ \sqrt{2} - 1}$

$\rm :\longmapsto\:\dfrac{\sqrt{x}}{\sqrt{y} } = \dfrac{ \sqrt{2} + 1}{ \sqrt{2} - 1}$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{x}{y} = \dfrac{ {( \sqrt{2} + 1)}^{2} }{ {( \sqrt{2} - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{x}{y} = \dfrac{2 + 1 + 2 \sqrt{2} }{2 + 1 - 2 \sqrt{2} }$

$\rm \implies\:\boxed{ \tt{ \: \frac{x}{y} = \frac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} } \: }}$

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1. If x and y are two positive real numbers, then

Arithmetic Mean

$\boxed{ \tt{ \: A = \frac{x +y }{2} \: }}$

Geometric mean (G)

$\boxed{ \tt{ \: G = \sqrt{xy} \: }}$

3. Harmonic mean (H)

$\boxed{ \tt{ \: H = \frac{2xy}{x + y}}}$

2. Between any two distinct positive real numbers then

$\boxed{ \tt{ \: {G}^{2} = A H \: }}$

and

$\boxed{ \tt{ \: A > G > H \: }}$