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Question

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


A
(4,6)
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B
(6,4)
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C
(4,6)
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D
(4,6)
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Solution

The correct option is D $$(6,-4)$$
If $$\alpha ,\beta ,\gamma ,\delta$$ be the four $$+ive$$ roots then
$$\dfrac { \alpha +\beta +\gamma +\delta  }{ 4 } =\dfrac { 4 }{ 4 } ={ (\alpha \beta \gamma \delta ) }^{ 1/4 }$$
Above shows that $$A.M$$. Of the four roots is equal to their $$G.M$$. 
Hence $$\alpha ,\beta ,\gamma ,\delta$$ are all equal and each equal to $$1$$.
$$\therefore \quad E={ (x-1) }^{ 4 }={ x }^{ 4 }-{ 4x }^{ 3 }+{ 6x }^{ 2 }-4x+1$$
Comparing, $$a=6,b=-4$$.

Mathematics

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