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(D) Let the four real roots be $\alpha, \beta, \gamma, \delta$. The equation is $(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0$.Comparing this w

Vedclass · Accepted Answer

$a = 6, b = -4$

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(D) Let the four real roots be $\alpha, \beta, \gamma, \delta$. The equation is $(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0$.Comparing this with $x^4 - 4x^3 + ax^2 + bx + 1 = 0$,we get:$\sum \alpha = 4$,$\sum \alpha\beta = a$,$\sum \alpha\beta\gamma = -b$,and $\alpha\beta\gamma\delta = 1$.For real positive roots,the Arithmetic Mean $(AM)$ $\ge$ Geometric Mean $(GM)$.$AM = \frac{\alpha + \beta + \gamma + \delta}{4} = \frac{4}{4} = 1$.$GM = (\alpha\beta\gamma\delta)^{1/4} = (1)^{1/4} = 1$.Since $AM = GM = 1$,all roots must be equal,i.e.,$\alpha = \beta = \gamma = \delta = 1$.Now,$a = \sum \alpha\beta = \binom{4}{2} 	imes (1 	imes 1) = 6 	imes 1 = 6$.$-b = \sum \alpha\beta\gamma = \binom{4}{3} 	imes (1 	imes 1 	imes 1) = 4 	imes 1 = 4$,so $b = -4$.Thus,$a = 6$ and $b = -4$.

The values of $a$ and $b$ for which the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots are:

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