Question

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

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

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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