Let two numbers have an arithmetic mean of 9 | vedclass

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(B) Let the two numbers be $x_1$ and $x_2$.The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.The geometric mean

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$x^2 - 18x + 16 = 0$

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(B) Let the two numbers be $x_1$ and $x_2$.The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.The geometric mean is given by $\sqrt{x_1 x_2} = 4$,which implies $x_1 x_2 = 4^2 = 16$.$A$ quadratic equation with roots $x_1$ and $x_2$ is given by the formula: $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - 18x + 16 = 0$.

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

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