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(B) Given the quadratic equation: ${x^2} - 2Ax + {G^2} = 0$  $(i)$Let $a$ and $b$ be two positive numbers such that their arithmetic mean is $A$ and g

Vedclass · Accepted Answer

$A > G$

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(B) Given the quadratic equation: ${x^2} - 2Ax + {G^2} = 0$  $(i)$Let $a$ and $b$ be two positive numbers such that their arithmetic mean is $A$ and geometric mean is $G$.Then,$A = \frac{a + b}{2}$ and $G^2 = ab$.Substituting these into equation $(i)$,we get: ${x^2} - (a + b)x + ab = 0$.Factoring the quadratic equation: ${x^2} - ax - bx + ab = 0$ $\Rightarrow x(x - a) - b(x - a) = 0$ $\Rightarrow (x - a)(x - b) = 0$.Thus,the roots of the equation are $a$ and $b$.For real roots $a$ and $b$ to exist,the discriminant $D$ of the quadratic equation must be non-negative: $D = (-2A)^2 - 4(1)(G^2) \ge 0$.$4A^2 - 4G^2 \ge 0 \Rightarrow A^2 \ge G^2$.Since $A$ and $G$ are means of positive numbers,$A, G > 0$,so $A \ge G$.Specifically,$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{1}{2}(\sqrt{a} - \sqrt{b})^2 \ge 0$.Therefore,$A \ge G$.

If $A$ and $G$ are arithmetic and geometric means and ${x^2} - 2Ax + {G^2} = 0$,then

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