Let `a ,b ,c` be real. If `a x^2+b x+c=0` has two real roots `alphaa n dbeta,w h e r ealpha<<-1a n dbeta>>1` , then show that `1+c/a+|b/a|<0`

Options

A. `lt 0`
B. `gt 0`
C. `le 0`
D. none of these

Answer

A

Explanation

Let `f(x) = ax^(2) + bx + c`. Clearly, -1 and 1 lie between the roots of f(x) = 0.
`therefore" "a f(1) lt 0 and a f(-1) lt 0`
`rArr" "a(a+b+c) lt 0 and a(a-b-c) lt 0`
`rArr" "a^(2)(1+(b)/(a)+(c)/(a)) lt 0 and a^(2) (1-(b)/(a)+(c)/(a)) lt 0`
`rArr" "1+(b)/(a)+(c)/(a) lt 0 and 1 -(b)/(a)+(c)/(a) lt 0 rArr 1+|(b)/(a)|+(c)/(a) lt 0`