Question

the sum of two numbers is 6 times their geometry mean . show that numbers are in the ration (3+2√2) : (3-2√2)​

Answer 1

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

Let assume that x and y are two positive numbers such that x > y and according to statement, sum of these two numbers is 6 times their geometric mean.

So,

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

can be further rewritten as

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

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

can be further rewritten as

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

On applying Componendo and Dividendo, we get

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

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

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

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

can be further rewritten as

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

On applying Componendo and Dividendo, we get

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

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

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

On squaring both sides, we get

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

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

$\rm\implies \:\rm \: \dfrac{x}{y} = \dfrac{3 + 2 \sqrt{2} }{3- 2 \sqrt{2} }$

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FORMULA USED

If x and y are two positive real numbers, then geometric mean (GM) is given by

$\boxed{\tt{ \: \rm \: \: GM \: = \: \sqrt{xy} \: \: }}$

If a : b :: c : d, then Componendo and Dividendo is

$\boxed{\tt{ \rm \: \: \: \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: \: }}$

$\boxed{\tt{ \rm \: \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }}$

$\boxed{\tt{ \rm \: \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: \: }}$

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ADDITIONAL INFORMATION

If x and y are two positive real numbers, then Arithmetic Mean (AM) is given by

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

If x and y are two positive real numbers, then Harmonic mean (HM) is given by

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

Relationship between Arithmetic mean, Geometric mean and Harmonic mean.

$\boxed{\tt{ \: \: AM \: \geqslant \: GM \: \geqslant \: HM \: }}$

$\boxed{\tt{ \: \: {GM}^{2} = AM \times HM \: \: }}$

$\boxed{\tt{ \: \: AM, \: GM, \: HM \: are \: in \: geometric \: progression \: }}$

Answer 2

Consider :

Assume that the two numbers can be a and b

We know that,

The geometric mean between a and b is √(a×b).

So ,

➡ Geometric Mean = √ab

Given Condition :

• The sum of two numbers is 6 times their geometric mean .

According to given condition ,

$\rm \implies \: a + b = 6 \sqrt{ab} \: \: \: \: \: \: (1)$

By squaring both the sides we get ,

$\rm \implies \: (a + {b)}^{2} = 36ab$

Also,

$\rm \implies(a - {b)}^{2} = (a + {b)}^{2} - 4ab = 36ab - 4ab = 32ab$

So,

$\rm \implies \: a - b = (\sqrt{32} )( \sqrt{ab} )$

$\rm \implies \: a - b = 4 \sqrt{2} . \sqrt{ab} \: \: (2)$

By solving equation (1) and (2) we get ,

$\rm \implies \: \dfrac{a + b}{a - b} = \dfrac{6 \: \cancel{ \sqrt{ab}} }{4 \sqrt{2} \: . \: \cancel{ \sqrt{ab}}}$

$\rm \implies \: \dfrac{a + b}{a - b} = \dfrac{3}{2 \sqrt{2} }$

By applying componendno and dividendo we get ,

$\rm \implies \: \dfrac{(a + b) + (a - b)}{(a + b) - (a - b)} = \dfrac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

$\rm \implies \: \dfrac{2a}{2b} = \dfrac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

$\rm \implies \: \dfrac{a}{b} = \dfrac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

$\rm \: \therefore \: Thus \: required \: Ratio \: is \: (3 + 2 \sqrt{2}) : (3 -2 \sqrt{2})$