simple... but just a thread..!!

http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

let one root be imaginary and the other be real .

thus sum of roots will be imaginary = -b/a

but b and a are real numbers.

contradiction.

Look at my question first
the term 'conjugate pairs' has been used

u 2 proved that the other root will be imaginary...

as if a root is (2 + 3i) and the other is (-3i + 8). They too satisfy that Z1 + Z2 = real but r not in conjugate pairs

Here z1 + z2 = Real and at the same time z1.z2 = Real which is possible only when they occur in conjugate pairs...!!

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well one other simple way could hv been

x = -b ± √b² - 4ac2a

now, if D < 0 or that b² - 4ac < 0 where a,b,c E R

or, x = -b2a ± √b² - 4ac2a

which shows if they are imaginary then will occur in conjugate paris...