Question

The ratio of two positive real numbers is `p+sqrt(p^2-q^2):p-sqrt(p^2-q^2)`then find their ratio of AM and GM?

Answer 1

`p/q`.

Explanation:

Let the nos. be `x and y," where, x,y"in RR^+`.

By what is given, `x:y=(p+sqrt(p^2-q^2)):(p-sqrt(p^2-q^2))`.

`:. x/(p+sqrt(p^2-q^2))=y/(p-sqrt(p^2-q^2))=lambda," say"`.

`:. x=lambda(p+sqrt(p^2-q^2)) and y=lambda(p-sqrt(p^2-q^2))`.

Now, the AM `A` of `x,y` is, `A=(x+y)/2=lambdap`, and, their

GM `G=sqrt(xy)=sqrt[lambda^2{p^2-(p^2-q^2)}]=lambdaq`.

Clearly, `"the desired ratio"=A/G=(lambdap)/(lambdaq)=p/q`.

Answer 2

`p/q`

Explanation:

I am going to use the same notation as in this answer . In fact there is no real necessity of this solution (as the problem has already been solved quite nicely) - except that it illustrates the use of a technique I love very much!

According to the problem

`x/y = (p + sqrt(p^2-q^2))/(p - sqrt (p^2-q^2))`

Using componendo and dividendo (this is the favorite technique I alluded to above) we get

`(x+y)/(x-y) = p/sqrt(p^2-q^2) implies`

`((x+y)/(x-y))^2 = p^2/(p^2-q^2) implies`

`(x+y)^2/((x+y)^2-(x-y)^2) = p^2/(p^2-(p^2-q^2)) implies`

`(x+y)^2/(4xy) = p^2/q^2 implies`

`(x+y)/(2sqrt(xy)) = p/q`

• which is the required AM : GM ratio.