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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.

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  • $\begingroup$ 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
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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
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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.

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Well, the solutions to the equation are

$$x=\frac{2+k}{2}\pm\frac12\sqrt{(-2-k)^2-4\cdot1\cdot-3}$$

or, simplified,

$$x=\frac{2+k}2\pm\frac12\sqrt{(2+k)^2+12}$$

and this is equal to

$$x=\frac{2+k}2\pm\sqrt{\left(\frac{2+k}2\right)^2+3}$$

The important step here is

$$\sqrt{\left(\frac{2+k}2\right)^2+3}>\sqrt{\left(\frac{2+k}2\right)^2}=\left|\frac{2+k}2\right|$$

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

$$\frac{2+k}2+\sqrt{\left(\frac{2+k}2\right)^2+3}>\frac{2+k}2+\left|\frac{2+k}2\right|\geq0$$

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

$$\frac{2+k}2-\sqrt{\left(\frac{2+k}2\right)^2+3}<\frac{2+k}2-\left|\frac{2+k}2\right|\leq0$$

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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