If A and G represent the arithmetic mean and | vedclass

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(B) Let the two numbers be $a$ and $b$. The arithmetic mean is $A = \frac{a + b}{2}$ and the geometric mean is $G = \sqrt{ab}$.Thus,$G^2 = ab$.The giv

Vedclass · Accepted Answer

$A > G$

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(B) Let the two numbers be $a$ and $b$. The arithmetic mean is $A = \frac{a + b}{2}$ and the geometric mean is $G = \sqrt{ab}$.Thus,$G^2 = ab$.The given equation is $x^2 - 2Ax + G^2 = 0$.Substituting $2A = a + b$ and $G^2 = ab$,we get $x^2 - (a + b)x + ab = 0$.Factoring the quadratic equation: $(x - a)(x - b) = 0$.This implies that $a$ and $b$ are the roots of the equation.For any two positive real numbers $a$ and $b$,the arithmetic mean is always greater than or equal to the geometric mean $(A \ge G)$.Specifically,$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$.Since $a$ and $b$ are roots of a quadratic equation with real coefficients,$A \ge G$ holds true.

If $A$ and $G$ represent the arithmetic mean and geometric mean respectively,and $x^2 - 2Ax + G^2 = 0$,then which of the following is true?

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