If (m – 1) a 12 – 2m a 2 < 0, then prove that x m + a 1 x m – 1 + a 2 x m – 2 + ..... + a m–1 x + a 0 = 0 has at least one non real root (a 1 , a 2 , ....., a m ∈ R)
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Let f(x) = x m + a 1 x m – 1 + a 2 x m – 1 + ...... + a 0 if possible, let f(x) = 0 has 'm' real roots, then by Roll's thearem, f '(x) = 0 must have "(m – 1)" real roots, f ''(x) = 0 must have "(m – 2)" real roots and so on, f m – 2 (x) = 0 must have 2 real roots, 
x 2 + a 1 (m – 1)! x + a 2 (m – 2)! = 0 must have 2 real roots or 
x 2 + a 1 (m – 1) + a 2 = 0 must have 2 real roots
D = a 12 (m – 1) 2 – 2m (m – 1) a 2
= (m – 1) [(m – 1) a 12 – 2a 2 ]
which is –ve, so our
allumption is wrong. Hence proved.
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