Let the equation $ax^2+bx+c=0$ have the roots $\alpha$ and $\beta$, then what is $\alpha-\beta$ in terms of $a$, $b$, and $c$?

Well, we may write $$(\alpha-\beta)^2=(\alpha+\beta)^2 -4\alpha \beta$$ $$\mbox{or}\ (\alpha-\beta)^2= \left(-\frac{b}{a}\right)^2-4\frac{c}{a}$$ $$\mbox{or}\ (\alpha-\beta)^2 = \frac{b^2}{a^2}-4\frac{c}{a}$$ $$\mbox{or}\ (\alpha-\beta)^2=\frac{b^2-4ac}{a^2}$$ $$\mbox{or}\ \alpha-\beta= \pm\sqrt{\frac{b^2-4ac}{a^2}} $$ $$\mbox{or}\ \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}.$$

Now, I know that the true value of $\alpha-\beta$ is $\frac{\sqrt{b^2-4ac}}{a}$, but what about $-\frac{\sqrt{b^2-4ac}}{a}$? Should I really get this other value or have I made some mistake or should I ignore the other value?

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    $\begingroup$ It just depends what order you subtract the roots in. $\endgroup$ – paw88789 Jan 28 '15 at 15:51

Who confirms the true value?

If $\alpha,\beta$ are real,

If $\alpha-\beta\ge0\iff\alpha\ge\beta ,\alpha-\beta=+\dfrac{\sqrt{b^2-4ca}}{|a|}$

Else $\alpha<\beta\implies\alpha-\beta=-\dfrac{\sqrt{b^2-4ca}}{|a|}$

If at least one of $\alpha,\beta$ is complex see

  1. http://www.cut-the-knot.org/do_you_know/complex_compare.shtml

  2. Total ordering on complex numbers

  3. Comparing complex numbers

  • $\begingroup$ i.e. $\pm \frac{\sqrt{b^2-4ac}}{a}$ cannot both be true at the same time? $\endgroup$ – Samama Fahim Jan 28 '15 at 16:05
  • $\begingroup$ @SamamaFahim, When we will have $x=-x$ $\endgroup$ – lab bhattacharjee Jan 28 '15 at 16:06
  • $\begingroup$ So we should not just write $\alpha-\beta = \frac{\sqrt{b^2-4ac}}{a}$, we should also add that $\alpha \geq \beta$, shouldn't we? $\endgroup$ – Samama Fahim Jan 28 '15 at 16:11
  • $\begingroup$ @SamamaFahim, The last part is the condition and the other condition as in the answer :$\alpha,\beta$ must be real $\endgroup$ – lab bhattacharjee Jan 28 '15 at 16:13
  • $\begingroup$ What I mean is that in the solution they just pick the positive value without stating that they supposed $\alpha \geq \beta$. This may confuse the reader. $\endgroup$ – Samama Fahim Jan 28 '15 at 16:17

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