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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 .

paradox xyz cool , 14 Years ago
Grade 11
anser 1 Answers
mycroft holmes

Last Activity: 14 Years ago

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

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