Equation `x^4ax^3+bx^2+cx+1=0` has real roots (a,b,c are non-negative).
Minimum non-negative real value of b is

Options

A. 12
B. 15
C. 6
D. 10

Answer

C

Explanation

Since, a,b,c `ge` 0 roots must be negative. Let the roots be
`x_(1), x_(2),x_(3),x_(4) (lt 0)`. Then.
`x_(1) + x_(2) + x_(3) + x_(4) = -a`
`x_(1)x_(2) + x_(1)x_(3) + x_(1)x_(4) + x_(2)x_(3) + x_(2)x_(4) + x_(3)x_(4) = b`
`x_(1)x_(2)x_(3) + x_(2)x_(3)x_(4) + x_(3)x_(4)x_(1) + x_(3)x_(4) x_(1) + x_(4)x_(1)x_(2) = - c`
`x_(1)x_(2)x_(3)x_(4) = 1`
Now, `((-x_(1)) + (-x_(2)) + (-x_(3)) + (-x_(4)))/(4) ge [(-x_(1))(-x_(2))(-x_(3))(-x_(4))]^(1//4)`
`( :' A.M ge G.M)`
`implies (a)/(4) ge 1`
Hence, the minimum value of a is 4. similarly,
`(x_(1)x_(2) + x_(1) x_(3) + ..... + x_(3)x_(4))/(6) ge [x_(1)^(3) x_(2)^(3) x_(3)^(3) x_(4)^(4)]^(1//4)`
`implies (b)/(6) ge 1`
or `b ge 6`
Hence, the minimum value of is 6. Finally
`(x_(1)x_(2)x_(3) - x_(2)x_(3)x_(4) - x_(3)x_(4)x_(1) - x_(4)x_(1)x_(2))/(4) ge [x_(1)^(3) x_(2)^(3) x_(3)^(3) x_(4)^(3)]^(1//4)`
`implies (c )/(4) ge 1`
or c ge 4`
Hence, the minimum value of c is 4.