Let a, b, c be real. If ax² + bx + c = 0 has two real roots α and β , where α < -1 and β > 1, then show that 1 + c/a + (modulus of b/a) < 0 .

WLOG we may assume b/a>0 [ otherwise consider the quadratic with roots -α, and -β, notice the conditions remain identical]

Now, 1 + b/a + c/a = 1 + α+β+αβ = (1+α)(1+β).

Now α<-1, so 1 + α<0. Also β>1, so 1+β>0. Hence (1+α)(1+β)<0

Hence 1 + b/a + c/a<0