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Multiple Choice:

If $c$ is a positive real number, then the equation $2x^2 - 3x - c = 0$ has:

(a) No Solutions (b) one solution (c) two solutions (d) three solutions


Can we assume $c$ to be any positive real number (for e.g. 6) and then use quadratic formula to find the solutions?

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It has always two solutions, over the complex numbers. It has never three solutions over any field. – Dietrich Burde Dec 5 '13 at 21:41
The problem might (and probably is ) for the real numbers. Also, there might be an issue of multiplicity. – LASV Dec 5 '13 at 21:43
It surely is for the real numbers. But one should think about it. – Dietrich Burde Dec 5 '13 at 21:43
up vote 2 down vote accepted

$D=(-3)^2-4\cdot 2\cdot (-c)=9+8c>0$. So we have two solutions

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No, $c$ is an arbitrary positive number, so do NOT fix it. Yes, use the quadratic equation. In particular, use the discriminant.

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Yes, in this particular question, your attempt will give the right answer!

The question assumes implicitly that the correct answer for all $c\in \mathbb{R}$ is the same choice from (a), (b), (c), (d). So, if the equation has, say, $n$ roots for $c=6$, then (given that the question is correct!) it has $n$ roots for all $c\in \mathbb{R}$.

I should add, this question does not seem a good one to me, I mean the choices are not designed good enough. For example one of the choices could be (d) (a) and (b) are both possible.

P.S. This (using a particular example when we know that the answer is the same for all examples) is a good trick! It is also mentioned in one of the chapters in Martin Gardner's beautiful book, Aha! insight .

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I don't think it is a good way to do that. The question might have been wrong. – LASV Dec 5 '13 at 21:50
Not wrong but dependable on $c$. – LASV Dec 5 '13 at 22:08

You don't even need to use the discriminant. $C$ is a positive number, so the constant term in the quadratic is negative.

If the y-intercept is negative and the leading coefficient of a quadratic is positive, then it must intersect the coordinate axes twice.

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