0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – donlan 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
1
$\begingroup$

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

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

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

$\endgroup$
  • 1
    $\begingroup$ Got it. I don't think I ever learned that! wow. thx $\endgroup$ – donlan 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.