If the equation x^4 - 4x^3 + ax^2 + bx + 1 = 0 has four positive...

Let the four positive roots be $\alpha, \beta, \gamma, \delta$. Sum of roots = $\alpha + \beta + \gamma + \delta = 4$. Product of roots = $\alpha\beta\gamma\delta = 1$. A.M. of roots = $\frac{4}{4} = 1$. G.M. of roots = $\sqrt[4]{1} = 1$. Since A.M. = G.M., all roots must be equal to 1. i.e., $\alpha = \beta = \gamma = \delta = 1$. So the equation is $(x - 1)^4 = 0$. Expanding $(x - 1)^4$ gives $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Comparing with $x^4 - 4x^3 + ax^2 + bx + 1 = 0$, we get $a = 6$ and $b = -4$. Therefore, $a + b = 6 - 4 = 2$.