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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