I think I just need some background.

I've got the following quadratic equation:

$$ 1 - x - 2x^2 = (1-2x)(1 + x) $$

But if I solve it with the quadratic equation, I get the roots:

$$ \frac{1 + \sqrt{9}}{-4} = -1, \frac{1-\sqrt{9}}{-4} = +1/2 $$

And, logically, I can't think of why I wouldn't write:

$$ (1 + x)(1/2 - x) = \frac{1}{2} - \frac{1}{2}x - x^2 = \frac{1}{2}(1 - x - 2x^2) $$

So I guess my question is:

What is the standard I should be holding to when I recreate the function from the roots, so that this kind of mistake doesn't happen? (do I always start with $(1-ax)$ and solve for $a$ when $x = \text{the root}$, for example?)

And...does it matter? Obviously, the second equation just looks different...the relationships are the same...but I'd like to be operating the 'regular' way...

  • $\begingroup$ The final equality is false: it should be $\frac12(1-x-2x^2)$. $\endgroup$ – Bernard Mar 25 '16 at 14:52
  • $\begingroup$ just edited it... $\endgroup$ – Chris Mar 25 '16 at 14:52
  • $\begingroup$ The polynomials $p(x)$ and $a\cdot p(x)$ have the same roots, for any nonzero constant $a$. $\endgroup$ – symplectomorphic Mar 25 '16 at 14:52
  • $\begingroup$ (Which means that knowing the roots alone does not determine the polynomial. You need another piece of data, such as another point the polynomial passes through.) $\endgroup$ – symplectomorphic Mar 25 '16 at 14:53

If $\;\alpha,\,\beta\;$ are the roots of the quadratic $\;y=ax^2+bx+c\;$ , then you get


I think you just forgot the higher coefficient $\;a\;$ . You may want to google Vieta's Formulas

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  • 1
    $\begingroup$ Got it. I don't think I ever learned that! wow. thx $\endgroup$ – Chris Mar 25 '16 at 14:54
  • $\begingroup$ @bordeo It's very easy to prove, but you may want first to work out some simple examples. $\endgroup$ – DonAntonio Mar 25 '16 at 14:56

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