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(C) Let the two numbers be $a$ and $b$.Given that the arithmetic mean $A = \frac{a + b}{2}$,so $a + b = 2A$.Given that the geometric mean $G = \sqrt{a

Vedclass · Accepted Answer

$A \pm \sqrt{(A + G)(A - G)}$

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(C) Let the two numbers be $a$ and $b$.Given that the arithmetic mean $A = \frac{a + b}{2}$,so $a + b = 2A$.Given that the geometric mean $G = \sqrt{ab}$,so $ab = G^2$.The quadratic equation whose roots are $a$ and $b$ is given by $x^2 - (a + b)x + ab = 0$.Substituting the values,we get $x^2 - 2Ax + G^2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:$x = \frac{2A \pm \sqrt{(2A)^2 - 4G^2}}{2} = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = \frac{2A \pm 2\sqrt{A^2 - G^2}}{2} = A \pm \sqrt{A^2 - G^2}$.Since $A^2 - G^2 = (A + G)(A - G)$,the numbers are $A \pm \sqrt{(A + G)(A - G)}$.

If the arithmetic mean of two numbers is $A$ and the geometric mean is $G$,then the numbers are

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