If roots of the equation
$x^{4} - 8 x^{3} + b x^{2} + c x + 16 = 0$
are positive, then

b=8

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c=-32

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b=24

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c=32

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Solution

The correct option is C
b=24

If $x_{1}, x_{2}, x_{3}, x_{4}$ are four roots of (1), then
$x_{1} + x_{2} + x_{3} + x_{4} = 8 a n d x_{1} x_{2} x_{3} x_{4} = 16$
As A.M. $\geq$ G.M., we get
$\frac{1}{4} (x_{1} + x_{2} + x_{3} + x_{4}) \geq (x_{1} x_{2} x_{3} x_{4})^{1 / 4} = 2 i . e . A . M . = G . M . T h u s, x_{1} = x_{2} = x_{3} = x_{4} = 2 ∴ x^{4} - 8 x^{3} + b x^{2} + c x + 16 = (x - 2)^{4} \Rightarrow b = 24 a n d c = - 32$

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