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Here's the question I have. I don't need help with the calculation for I have that already.

Construct a (unique) quadratic equation for which the sum of the roots is 3 and the sum of the cubes of the roots is 63.

I know what roots satisfy this, yet isn't it true that $$f(x)=k(x-\alpha)(x-\beta)$$ holds under these conditions and thus the quadratic is NOT unique? After all I 'm looking for a specific quadratic but only need the roots satisfy the above conditions. Yet this holds $\forall{k}$, right?

Eleven-Eleven
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  • so uniqueness as mentioned in the problem is ONLY for the $\alpha$ and $\beta$? Doesn't the tone of the problem seem to indicate there is a unique quadratic equation that satisfy the condition on the roots, not the uniqueness of the roots? – Eleven-Eleven Apr 23 '13 at 20:00

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You want to solve $$\alpha + \beta = 3$$ $$\alpha^3 + \beta^3 = 63$$

So yes, if the solution for $(\alpha, \beta)$ is unique, then any quadratic of the form $$p(x) = k(x -\alpha )(x - \beta)$$ will satisfy the given criteria.

So, with respect to your question, the criteria is satisfied for any $p(x)$ where $k \neq 0$. One could say that the quadratic is unique up to a constant, nonzero factor, i.e., for any given nonzero scalar k, the corresponding quadratic $p(x)$ is unique.

amWhy
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  • So it is unique with regard to a scalar k. – Eleven-Eleven Apr 23 '13 at 20:06
  • Yes...the criteria is satisfied for any $k \neq 0$. So, one could say that the quadratic is unique up to a constant, nonzero factor. Yes, exactly. I think you could simply say that for any given nonzero scalar k, the corresponding quadratic is unique. – amWhy Apr 23 '13 at 20:06
  • That was what i was looking for. I already had that idea, but I didn't know if I was missing something to solve for k. But thank you for verifying my question. – Eleven-Eleven Apr 23 '13 at 20:08
  • You're welcome. It just took me time to type the rest out...you had good intuition regarding the problem! – amWhy Apr 23 '13 at 20:09
  • Thanks @amWhy. My professor is quite rigorous and you can definitely get points off for stuff like that. can you amend your answer as to not include the answer of (x,y) as it is active homework? – Eleven-Eleven Apr 23 '13 at 20:10
  • Is it possible that the question asked for a monic polynomial? In this case it's unique. – vadim123 Apr 23 '13 at 20:28
  • @amWhy: Nice response and follow-ups! +1 – Amzoti Apr 24 '13 at 00:48
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    @vadim123, the problem is worded exactly as my professor wrote it, even down to the (unique) in parentheses. Perhaps this is the giveaway that it's "unique to an extent", the extent here being that it holds for a constant $k\neq{0}$. – Eleven-Eleven Apr 24 '13 at 03:21