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Question
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3 + 2sqrt2) ":" (3 - 2sqrt2)`.
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Solution
Let the two numbers be a and b.
geometric mean of a and b = `sqrt"ab"`
Given: a + b = `6sqrt"ab"`
`"a"+ "b" + 2sqrt"ab" = 8sqrt"ab"`
`(sqrt"a" + sqrt"b")^2 = 8sqrt"ab"` .......(i)
`"a" + "b" - 2 sqrt"ab" = 4sqrt"ab"`
`(sqrt"a" - sqrt"b")^2 = 4sqrt"ab"` .......(ii)
Dividing equation (i) by (ii), we get
`(sqrt"a" + sqrt"b")^2/(sqrt"a" - sqrt"b")^2 = (8sqrt"ab")/(4sqrt"ab") = 2`
or `(sqrt"a" + sqrt"b")/(sqrt"a" - sqrt"b") = sqrt2/1`
⇒ `((sqrt"a" + sqrt"b") + (sqrt"a" - sqrt"b"))/((sqrt"a" + sqrt"b") - (sqrt"a" - sqrt"b")) = (sqrt2 + 1)/(sqrt2 - 1)`
`(2sqrt"a")/(2sqrt"b") = sqrt"a"/sqrt"b" = (sqrt2 + 1)/(sqrt2 - 1)`
On squaring, `"a"/"b" =(sqrt2 + 1)^2/(sqrt2 - 1)^2 = (3 + 2sqrt2)/(3 - 2sqrt2)`
Hence, `"a"/"b" =(3 + 2sqrt2)/(3 - 2sqrt2)`
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