If $A$ and $G$ are the $A.M.$ and $G.M.$ respectively between two positive numbers,prove that the numbers are $A \pm \sqrt{(A + G)(A - G)}$.

It is given that $A$ and $G$ are the $A.M.$ and $G.M.$ between two positive numbers.
Let these two positive numbers be $a$ and $b$.
$\therefore A.M. = A = \frac{a+b}{2}$ .........$(1)$
$G.M. = G = \sqrt{ab}$ .........$(2)$
From $(1)$ and $(2)$,we obtain:
$a+b = 2A$ .........$(3)$
$ab = G^2$ .........$(4)$
We know the identity $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the values from $(3)$ and $(4)$:
$(a-b)^2 = (2A)^2 - 4G^2 = 4A^2 - 4G^2 = 4(A^2 - G^2)$
$(a-b)^2 = 4(A+G)(A-G)$
$(a-b) = 2\sqrt{(A+G)(A-G)}$ .........$(5)$
Adding $(3)$ and $(5)$:
$2a = 2A + 2\sqrt{(A+G)(A-G)} \Rightarrow a = A + \sqrt{(A+G)(A-G)}$
Subtracting $(5)$ from $(3)$:
$2b = 2A - 2\sqrt{(A+G)(A-G)} \Rightarrow b = A - \sqrt{(A+G)(A-G)}$
Thus,the two numbers are $A \pm \sqrt{(A+G)(A-G)}$.