If `a`, `b`, `c` are positive numbers such that `a gt b gt c` and the equation `(a+b-2c)x^(2)+(b+c-2a)x+(c+a-2b)=0` has a root in the interval `(-1,0)`, then

Options

A. `c+alt2b`
B. both roots are rational
C. the equation `ax^(2)+2bx+c=0` have both negative real roots
D. the equation `cx^(2)+2ax+b=0` have both positive roots

Answer

D

Explanation

If `a, b, c` are the ………..
`f(-1)f(0)lt 0`
`(2a-b-c)(c+a-2b)lt 0`
`((a-b)+(a-c))(c+a-2b)lt 0`
Since `f(1) = 0` one root is 1 other : `(c+a-2b)/(a+b-2c)`
using `(c+a-2b)` both roots of `{:(ax^(2)+2bx+c=0),(cx^(2)+2ax+b=0):}}` are
real and negative
`alpham, beta = (-2b+- sqrt(4b^(2)-4ac))/(2a)`
`= (-b+- sqrt(b^(2)-4ac))/(2a)`
`{:(=(-b+-sqrt(b^(2)-4ac))/(a)),(= -b+- ("less than b")):}}-ve`