The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.

Answer

N/a

Explanation

Let the numbers are 'a' and 'b'
Given that, `a+b=6sqrt(ab)`
`implies (a+b)/(2sqrt(ab))=(3)/(1)`
From componendo/dividendo rule
`(a+b+2sqrt(ab))/(a+b-2sqrt(ab))=(3+1)/(3-1)`
`implies ((sqrt(a)+sqrt(b))^(2))/((sqrt(a)-sqrt(b))^(2))=(2)/(1)`
`(sqrt(a)+sqrt(b))/(sqrt(a)-sqrt(b))=(sqrt(2))/(1)`
Again from componendo/dividendo rule
`((sqrt(a)+sqrt(b))+(sqrt(a)-sqrt(b)))/((sqrt(a)+sqrt(b))-(sqrt(a)-sqrt(b)))=(sqrt(2)+1)/(sqrt(2)-1)`
`implies (2sqrt(a))/(2sqrtb)=(sqrt(2)+1)/(sqrt(2)-1)`
`= (a)/(b)=((sqrt(2)+1)^(2))/((sqrt(2)-1)^(2))`
`implies (a)/(b)=(3+2sqrt(2))/(3-2sqrt(2))" "` Hence Proved.