Question

Answer It With explanation ?

`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]` .

Answer 1

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