If the equation $x^{4} - {4 x}^{3} + {a x}^{2} + b x + 1 = 0$ has four positive roots then $(a, b)$ is

(- 4, 6)

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(6, - 4)

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(- 4, - 6)

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(4, - 6)

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Solution

The correct option is D $(6, - 4)$
If $α, β, γ, δ$ be the four $+ i v e$ roots then
$\frac{α + β + γ + δ}{4} = \frac{4}{4} = {(α β γ δ)}^{1 / 4}$
Above shows that $A . M$ . Of the four roots is equal to their $G . M$ .
Hence $α, β, γ, δ$ are all equal and each equal to $1$ .
$∴ E = {(x - 1)}^{4} = x^{4} - {4 x}^{3} + {6 x}^{2} - 4 x + 1$
Comparing, $a = 6, b = - 4$ .

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x^{4} - 4 x^{3} + a x^{2} + b x + 1 = 0

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