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?

2 Answers
Jun 11, 2018

# 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#.

Jun 11, 2018

#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.