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#A# and #G# are the AM (Arithmetic Mean) and GM (Geometric Mean) of two positive numbers. Then prove that the two numbers are #A± √[A^2-G^2]# .

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Answer 1 · 2018-08-12T08:25:52

Let the two numbers be #x# & #y# #(x>y)#

If, #A# is their arithmetic mean,then we can write,

#2A=x+y.... 1#

And, #G# being their geometric mean,we can write,

#G^2=xy#

Now,

#(x-y)^2=(x+y)^2 -4xy#

#=(2A)^2 -4G^2#

#=4(A^2 - G^2)#

So, #(x-y)=2 sqrt(A^2-G^2).....2#

Solving #1# and #2# we get,

#x=A+sqrt(A^2-G^2)#

And #y=A-sqrt(A^2-G^2)#

Proved