3
$\begingroup$

Why, in the rules regarding the discriminant determining whether roots are real or not, does it matter that the quadratic has real coefficients?

For the roots to be real we have $$\Delta=b^2-4ac\ge 0 \implies b^2 \ge 4ac$$

If $b$ was imaginary $b^2$ would end up being negative. If the product $4ac < b^2$ we can still have $b^2 - 4ac < 0 $. In this scenario the rules regarding the discriminant does not hold.

$\endgroup$
7
  • 3
    $\begingroup$ If the coefficients are not real, and $b^2-4ac\gt 0$, you could still have no real roots because $b$ or $a$ are complex. For example, $x^2+ix-1$ has $b^2-4ac=-1+4=3\gt 0$, but it does not have real roots, as the roots are $\frac{\sqrt{3}-i}{2}$ and $\frac{-\sqrt{3}-i}{2}$. So even though the discriminant is positive, the roots are nonreal complex. $\endgroup$ Dec 6, 2021 at 20:17
  • 3
    $\begingroup$ It is not clear which "rules regarding the discriminant" the question is asking about. If the coefficients are not real then the discriminant will not be real in general, so you cannot compare it to $0$, or say it is positive or negative. $\endgroup$
    – dxiv
    Dec 6, 2021 at 22:20
  • $\begingroup$ @dxiv I see. That explains why Arturo's solution is the better/correct one. Concerning "rules regarding the discriminant", I've only being introduced to the discriminant advising if the solution to a quadratic is real, double root or not real. $\endgroup$
    – Edmund
    Dec 6, 2021 at 22:37
  • 1
    $\begingroup$ Given $\quad x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\quad $ the $b$ outside the radical makes the roots imaginary in any case. $\endgroup$
    – poetasis
    Dec 7, 2021 at 1:25
  • $\begingroup$ @poetasis But for example $\,ix^2-3ix+2i=0\,$ has real roots (and negative discriminant). $\endgroup$
    – dxiv
    Dec 7, 2021 at 4:03

1 Answer 1

3
$\begingroup$

Consider the quadratic $\,az^2+bz+c\,$ with complex coefficients $\,a,b,c \in \mathbb C\,$, $\,a \ne 0\,$.

  • The roots are still given by the same formula as in the real case $\,z_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}\,$ where $\,\Delta =b^2-4ac\,$ is the discriminant. In particular, $\,\Delta = 0\,$ iff the quadratic has a double root, and in that case $\,z_1=z_2=-\frac{b}{2a}\,$. This can be easily proved the same way as in the real case, for example by completing the square.

  • The discriminant $\,\Delta\,$ is complex in general, and complex numbers are not ordered, so $\,\Delta\,$ has no "sign" to speak of, and does not indicate the nature of the roots. Besides, unlike the real case, a quadratic with complex coefficients can have one real and one non-real complex root.

It is possible to determine the nature of the roots of a complex quadratic by reducing it to the case of two real quadratics. To simplify the calculations, let $\,z=x+iy\,$, $\,\frac{b}{a}=b'+ib''$, $\,\frac{c}{a}=c'+ic''$, then dividing the equation by $\,a \ne 0\,$ and isolating the real and imaginary parts gives:

$$ (x+iy)^2+(b'+ib'')(x+iy)+c'+ic'' = 0 $$

$$ \iff\;\;\;\; \begin{cases} \begin{align} x^2 - y^2+b'x-b''y + c' &= 0 \\ 2xy+b'y+b''x+c'' &= 0 \end{align} \end{cases} $$

For the equation to have a real root, the system must have a solution with $\,y=0\,$:

$$ \begin{cases} \begin{align} x^2 +b'x + c' &= 0 \\ b''x+c'' &= 0 \end{align} \end{cases} $$

  • If $\,b''=0\,$ the second equation requires $\,c''=0\,$, which is the case where $\frac{b}{a}, \frac{c}{a} \in \mathbb R$, so the original quadratic is equivalent to one with real coefficients. The roots will be either both complex, or both real, depending on the sign of $\,\Delta'=b'^{\,2}-4c'\,$.

  • If $\,b'' \ne 0\,$ then the second equation gives $\,x=-\frac{c''}{b''}\,$, and substituting back into the first equation $\,c''^{\,2}+b'b''c''+b''^{\,2}c'=0\,$, which is the condition for at least one real root.

$\endgroup$
0

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.