IIT JEE , AIEEE Forum - Mathematics , Algebra

eklavya2009

if ax² + bx + c =0 (a>0) has two roots

and

such that

<-2 and

> 2 then a + |b| + c <0 ..........(T/F)

sboosy

Since it is an upward drawn parabola ...the value of f for those values between the roots (-2 and 2 minimum) are negative

so f(1)<0 =>a+b+c<0

f(-1)<0 =>a-b+c<0

Thus a+|b|+c<0

amulye

may b dis is anoder medod

a+c+IbI = a[1+c/a +IbI/a]

=a[1+alpha x beta +I-(alpa+beta)I

= a[1-ve -ve]
=as a is +ve der4 d given is -ve