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 ?

#Algebra #MathProblem #Math

Easy Math Editor

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Comments

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.

C Lim - 7 years, 11 months ago

Could not follow your second argument. Please explain a bit further.

Nishant Sharma - 7 years, 11 months ago

What CL did is right. 1st prove that a,b,c are real. now assume a is not equal to zero. Divide the whole polynomial by a. Give the new constants a new name. We get f(x)=x^2+bx+c. As per on your condition if x is non real then f(x) should be non-real. Since c is real we can say that x^2+bx should be non real for non real x. But there exist some non-real x which makes x^2+bx real. hence not satisfying your condition. therefore we can say that there exist no nonzero a which satisfies your condition. Now consider a =0. We have p(x)=bx+c which is nonreal for every non real x. Hence Proved

Abhishek SETHI - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$