Question

The sum of two numbers is 6 times geometric means, show that numbers are in the ratio
$(3 + 2 \sqrt{2} ):(3 - 2 \sqrt{2} )$
​

Answer 1

$\sf \large \:\underline{ \red{Answer} }:-$

Let the two numbers be a and b.

$G.M. = \huge{ \sqrt{ab} }$

According to the Given condition,

$\bold{a + b = 6 \sqrt{ab} }$

$\bold{ = > (a + b {)}^{2} = 36(ab)...(1)}$

$\sf \large \:\underline{ \red{Also} },$

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

$= > \bold{a - b = \sqrt{32} \sqrt{ab} }$

$\bold{4 \sqrt{2} \sqrt{ab} \: \: \: ...(2) }$

Adding (1) and (2),

$\boxed{ \bold{ \red{we \: obtain}}} \sf{}$

$\bold{2a \: = (6 + 4 \sqrt{2}) \sqrt{ab} }$

$= > \bold{a = (3 + 2 \sqrt{2} ) \sqrt{ab} }$

$\boxed{Substituting \: the \: value \: of \: a \: in \: (1),}$

$\boxed{ \bold{ \red{we \: obtain}}}$

$\bold{b = 6 \sqrt{ab} - (3 + 2 \sqrt{2} ) \sqrt{ab} }$

$\bold{ = > b = (3 - 2 \sqrt{2} ) \sqrt{ab} }$

$\bold{ = > \bold{ \frac{a}{b}} = \huge{ \frac{(3 + 2 \sqrt{2} \sqrt{ab}) }{3 - 2 \sqrt{2} \sqrt{ab} } } \: = \: \frac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} } }$

$\sf \large \:\underline{ \red{Thus} },$

The required ratio is (3+2$\sqrt{2}$) : (3 - 2 $\sqrt{2}$)

Answer 2

Step-by-step explanation:

$\huge\underline\mathfrak{\green{Answer :-} }\:$

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