If the Equation X^4 - 4X^3+aX^2+bX+1 = 0 has four positive roots then find a, b?

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Solution

$x^{4} - 4 x^{3} + {ax}^{2} + bx + 1 = 0 Let x_{1}, x_{2}, x_{3}, x_{4} be the four positive roots . Now sum of roots = \frac{- (coefficient of x^{3})}{coefficient of x^{4}} = - \frac{- 4}{1} = 4 So x_{1} + x_{2} + x_{3} + x_{4} = 4 . . . . . . (1) Also product of roots = x_{1} x_{2} x_{3} x_{4} = \frac{constant term}{coefficient of x^{4}} = 1 Now we can write, \frac{1}{4} (x_{1} + x_{2} + x_{3} + x_{4}) = 1 = 1^{\frac{1}{4}} = {(x_{1} x_{2} x_{3} x_{4})}^{\frac{1}{4}} \frac{(x_{1} + x_{2} + x_{3} + x_{4})}{4} = {(x_{1} x_{2} x_{3} x_{4})}^{\frac{1}{4}} That is A . M = G . M . but that is possible if and only if x_{1} = x_{2} = x_{3} = x_{4} Now using equation 1, x_{1} = x_{2} = x_{3} = x_{4} = 1 It means all the roots are equal to 1 . SO the given equation becomes, {(x - 1)}^{4} = 0 {(x - 1)}^{2} {(x - 1)}^{2} = 0 (x^{2} - 2 x + 1) (x^{2} - 2 x + 1) = 0 x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0 Now comparing it with, x^{4} - 4 x^{3} + {ax}^{2} + bx + 1 = 0 we get a = 6 and b = - 4$
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