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# Polynomial problem

Let p(x)= $$a$$$$x^{2}$$ + $$b$$$$x$$ + $$c$$ be such that p(x) takes real values for real values of $$x$$ and non-real values for non-real values of $$x$$. Prove that $$a$$=0.

I tried it out taking $$a$$,$$b$$ and $$c$$ to be complex values and then implemented the conditions given in the problem which only got me to a weird expression. Any correct approach ?

Note by Nishant Sharma
4 years, 1 month ago

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Hint: first prove that a, b, c are real. For this it suffices to use the fact that p(-1), p(0), p(1) are real. Specifically, express a, b & c in terms of these three values.

Next, suppose a>0. Find a sufficiently small M<0 such that p(x) = M has no real roots, or equivalently, the discriminant for $$ax^2 + bx + (c-M)=0$$ is negative. Do something similar for a<0. · 4 years, 1 month ago

Could not follow your second argument. Please explain a bit further. · 4 years, 1 month ago