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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

$a = 6, b = -4$

Vedclass · Answer

(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:$\sum \alpha = 4, \sum \alpha\beta = a, \sum \alpha\beta\gamma = -b, \alpha\beta\gamma\delta = 1$.For real positive roots,by the $AM-GM$ inequality:$\frac{\alpha + \beta + \gamma + \delta}{4} \ge (\alpha\beta\gamma\delta)^{1/4}$.Substituting the values: $\frac{4}{4} \ge (1)^{1/4} \implies 1 \ge 1$.Since the $AM$ equals the $GM$,all roots must be equal: $\alpha = \beta = \gamma = \delta = 1$.Now,calculate $a$ and $b$:$a = \sum \alpha\beta = \binom{4}{2} 	imes (1 	imes 1) = 6 	imes 1 = 6$.$-b = \sum \alpha\beta\gamma = \binom{4}{3} 	imes (1 	imes 1 	imes 1) = 4 	imes 1 = 4 \implies b = -4$.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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