Prove that, for all values of $k$, the roots of the quadratic polynomial $x^2 - (2 + k) x - 3$ are real. Show further that the roots are of opposite signs.

For the first part I was able to demonstrate such by using the discriminant of the quadratic, then using the discriminant of the discriminant.

For the second part I was not able to demonstrate such.

  • $\begingroup$ For a polynomial of degree two, there are many ways to compute the roots. Which one do you use? What are the roots, if you leave $k$ in there as a variable? $\endgroup$ – Dirk Apr 11 '17 at 9:47
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    $\begingroup$ Do you know Vietas formulas? $\endgroup$ – kingW3 Apr 11 '17 at 9:49
  • $\begingroup$ From what I know you might either factorise or use formula although I don't see how this might help. $\endgroup$ – Gavish Apr 11 '17 at 9:51
  • $\begingroup$ Same for Vietas formula $\endgroup$ – Gavish Apr 11 '17 at 9:51
  • $\begingroup$ Maybe just write down the roots as you would compute them and then we can ask the question why (for all $k$) these two numbers will always have different signs. $\endgroup$ – Dirk Apr 11 '17 at 9:51

A general quadratic with roots $\alpha$ and $\beta$ can be written $$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$ The last, constant term is the product of the roots. In your case that equals $-3$. The roots must therefore have opposite signs.

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    $\begingroup$ I feel silly now. This is a much easier and more elegant solution (+1) $\endgroup$ – vrugtehagel Apr 11 '17 at 9:54
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    $\begingroup$ I think it's helpful and instructive to see multiple approaches and yours has value (+1) No need to feel silly. $\endgroup$ – PM. Apr 11 '17 at 9:56
  • $\begingroup$ The reason this works is because polynomials of different powers are linearly independent, right? $\endgroup$ – jpmc26 Apr 11 '17 at 11:36
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    $\begingroup$ @jpmc26 Perhaps this answers your question. The quadratic with roots alpha and beta is necessarily (x-alpha)(x-beta)=0. Expand this. $\endgroup$ – PM. Apr 11 '17 at 12:34
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    $\begingroup$ Your answer makes it sound like solving an easy Putnam problem.. It is about finding the right approach $\endgroup$ – Dennis Apr 11 '17 at 15:25

In the quadratic $P(x) = x^2 - (2 + k)x - 3 = 0$, the coefficient of $x^2$ is positive. Thus, $P(X)$ and $P(-X)$ is positive for some large enough $X>0$.

But $P(0) = -3$ is negative. So between $-X$ and $0$ there is a $x_1$ where $P(x_1) = 0$ and between $0$ and $X$ there is a $x_2$ where $P(x_2) = 0$: the polynomial has real roots and they are of opposite signs.


Well, the solutions to the equation are


or, simplified,


and this is equal to


The important step here is


and so if we choose $+$, then we see the root is positive:


When we choose $-$, we get something negative:


thus, the roots must be of different signs.

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    $\begingroup$ Or in general, the two values $-b \pm \sqrt{b^2 - 4ac}$ must have opposite signs iff $-4ac > 0$. $\endgroup$ – Peter Taylor Apr 11 '17 at 11:04
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    $\begingroup$ Which, in a sense, is equivalent to PM.'s answer. $\endgroup$ – vrugtehagel Apr 11 '17 at 11:13

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