The formula for finding the roots of a quadratic polynomial is as follows

$$x = \frac {-b \pm \sqrt{ b^2 - 4ac }}{2a} $$

what happens if you want to find the roots of a polynomial like this simplified one

$$ 3x^2 + x + 24 = 0 $$

then the square root value becomes

$$ \sqrt{ 1^2 - 4\cdot3\cdot24 } $$

$$ = \sqrt{ -287 } $$

which is the square root of a negative number, which isn't allowed. What do you do in this case? I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases or have i gone wrong somewhere along the path?

  • 1
    $\begingroup$ @Chandru1: Also, \cdot looks better than * don't you agree? $\endgroup$ Nov 11, 2010 at 10:00
  • $\begingroup$ The answers explain what is $\sqrt{-287}$. You can verify that $3x^{2}+x+24$ (like any other quadratic polynomial) can be factored as $3(x-x_{1})(x-x_{2}) $, even in this case where $x_{1,2}$ are non real roots, $3$ being the coefficient of $x^{2}$: $3\left( x+\frac{1}{6}+\frac{\sqrt{-287}}{6}\right) \left( x+\frac{1}{6}-\frac{-\sqrt{-287}}{6}\right) $ $=3\left( x+\frac{1}{6}\right) ^{2}-3\left( \frac{\sqrt{-287}}{6}\right) ^{2} $ $=3x^{2}+x+\frac{1}{12}-\frac{-287}{12}=3x^{2}+x+24$ $\endgroup$ Nov 11, 2010 at 13:02
  • $\begingroup$ Remember that the quadratic formula is just a generalization of completing the square, so it's really all the same. $\endgroup$
    – friedo
    Nov 11, 2010 at 15:58
  • $\begingroup$ In the last factor of my comment the minus signal before $\sqrt{-287}$ is a typo. $\endgroup$ Nov 12, 2010 at 10:24
  • 1
    $\begingroup$ @AD-StopPutin- Agreed. * is too computery. \cdot works well. Sometimes, \times works. $\endgroup$
    – badjohn
    May 29, 2023 at 11:23

3 Answers 3


The other answers are nice, so I won't reiterate them. It's worth pointing out, though, that when you say "I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases", you seem to be implying that these other methods might yield different results.

This isn't the case. If the quadratic formula fails to give you real-valued answers, then the other two methods (and every applicable method) will also fail, because the quadratic polynomial itself simply doesn't have any real-valued roots.

You can visualize this easily: as you may know, the graph of a quadratic polynomial is a parabola, and the real roots of a polynomial correspond to its graphs' zeros, i.e. points where the graph crosses the $x$-axis. So if you imagine a parabola with its vertex above the $x$-axis and opening upwards, you can see that it will never cross the $x$-axis -- it has no real zeros at all! alt graph of quadratic polynomial

This is exactly the situation with $y = 3x^2 + x + 24$ and every other quadratic polynomial for which $b^2 - 4ac$ (the discriminant) is negative.

Why isn't taking the square root of a negative number allowed?

Lastly (since it seems as though this may be one of your first introductions to complex numbers) let me clarify something that confuses my students greatly: don't let the words "real" and "imaginary" mislead you into thinking that real numbers are somehow genuine or actual and imaginary numbers are somehow fake.

Just as a "negative" number isn't sad, cynical, or irritated, and just as an "irrational" number hasn't lost its mind, "real" and "imaginary" are technical terms, and not meant to carry with them their English connotations. Although they earned those names due to prior misunderstandings about numbers, today we know that imaginary numbers are just as actual and valid as real numbers, and in fact complex (not "complicated") numbers have huge real-world applications and consequences.

Along these same lines: you say that taking the square root of a negative number "isn't allowed". More accurate would be to say that the square root of a negative number isn't a real number ("real", again, in the technical sense) because the square of a real number is always non-negative. If you require the result of taking the square root to be a real number, then you're right, taking the square root of a negative isn't allowed. But otherwise it's perfectly fine.

By analogy: you're allowed to divide the whole number 6 by the whole number 3, and the result is the whole number 2. But if you divide the whole number 1 by the whole number 2, then the result, $\frac{1}{2}$, isn't a whole number. If you require your operations on a certain class of numbers to produce a result from that same class of numbers, then lots of operations (like division, and subtraction too, for that matter) will have situations in which they aren't "allowed" to occur.

This requirement is sometimes very desirable, and there are certainly contexts in which you will want to impose it. But you should understand that imposing this requirement is a choice that you make because it fits the context of the problems you're working on. It's not divine law. If you find yourself in a situation in which you would like to take the square root of a negative number and produce the corresponding imaginary number, you are definitely "allowed" to do so.

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    $\begingroup$ I believe even Gauss himself lamented the poor choice of words $\endgroup$ Nov 11, 2010 at 12:32
  • $\begingroup$ Thanks Alex, your answer leads to a clear understanding of why the discriminant is negative, as well as why the method fails, which is actually more important to understanding the problem than the mechanics of working out the roots on the complex plane. Also a graph always helps. $\endgroup$ Nov 12, 2010 at 0:07

100% correct, and good observation.

To solve this, we define $\sqrt{-1}=i$ where $i$ is the imaginary unit

Then $\sqrt{-287}=\sqrt{287}i$, and we can solve as per the general quadratic formula. Numbers of the form $a+bi$ are known as complex numbers and are extremely useful.

In general the term $b^2-4ac$ is known as the discriminant of the quadratic equation. It should be clear that if $b^2-4ac>0$ there exists two real solutions, if $b^2-4ac=0$ there is one solution (the repeated root) and if $$b^2-4ac \lt 0$$ there are two complex solutions.

The quadratic formula is the most general way to solve the quadratic equation - so you are doing the right thing.

  • $\begingroup$ your explanation cuts off (i think the markup might be at fault), could you edit it so i can read the rest of it? It's good so far. $\endgroup$ Nov 11, 2010 at 9:24
  • $\begingroup$ @Aran Mulholland: this is a rare bug, did you try reloading the site? $\endgroup$ Nov 11, 2010 at 9:51
  • $\begingroup$ @Aran Mullholland: Yes, it doesn't appear to format correctly, but I believe it is. Hopefully, if it is not just a site bug, someone will edit it for me $\endgroup$
    – Juan S
    Nov 11, 2010 at 10:07
  • $\begingroup$ yep tried reloading a few times, the explanation cuts off "there is one solution (the repeated root) and if $b^2-4ac" (so you can see part of the markup is probably at fault as i assume this is mid sentence) - c'mon admins help us out! $\endgroup$ Nov 11, 2010 at 10:18

Yeah so have you read about complex numbers. The root will be $$x = \frac{-1 \pm{i} \sqrt{287}}{2 \times 3}$$ where $i=\sqrt{-1}$. Read more about complex numbers and when a polynomial can have complex roots, that happens when the *Discriminant* factor $b^{2}-4ac <0$.

  • $\begingroup$ so does that mean there is no general solution for solving polynomials, without straying into the complex plane? $\endgroup$ Nov 11, 2010 at 10:23
  • $\begingroup$ @Aran: Part of the reason for coming up with the concept of complex numbers was to be able to solve polynomial equations. And it gets crazier with cubic equations: the case where all roots are real requires in general radicals of complex numbers to express the solutions. $\endgroup$ Nov 11, 2010 at 10:33
  • $\begingroup$ Hmm, $\sqrt{-1} = i$, that feels a bit odd because the normal square root is not defined for negative numbers and the complex one is multivalued. $\endgroup$
    – JT_NL
    Nov 11, 2010 at 10:35
  • $\begingroup$ @Jonas: I prefer couching it as $i^2=-1$ just so we can sweep Riemann sheet arguments under the rug for a while (at least, up until the student gains the firepower to see what's going on). ;) $\endgroup$ Nov 11, 2010 at 10:40

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