Edit: I figured it out, I just miscalculated. No, there are no complex solutions to k.

For what values of $k$ does the equation $kx^2+3kx+k=-9-kx$ have real roots? Just to simplify it, I'll write the equation as $kx^2+4kx+k+9=0$. Now the obvious solution would probably be to plug it into the quadratic formula and try to solve such that the discriminant is greater than zero. In the quadratic formula, $a = k$, $b = 4k$, and $c = k + 9$. Plugging these values in, I get

$$\begin{align} &\frac{-4k±\sqrt{(4k)^2-4(k)(k+9)}}{2k}\\ &=\frac{-4k±\sqrt{16k^2-4k^2-36k}}{2k}\\ &=\frac{-4k±\sqrt{12k^2-36k}}{2k}\\ &=-2±\frac{\sqrt{12k^2-36k}}{2k}\\ &=-2±\frac{\sqrt{3k^2-9k}}{k} \end{align}$$

Since the only thing that matters is finding real values, $-2$ can just be ignored as well as the $\pm$ sign, since nothing really changes anyways. Now the logical thing to do here would be to take the discriminant, set it to be greater than or equal to zero, and get the inequality for all the real values of $k$ (in case you're wondering, $k ≤ 0$ and $k ≥ 3$). But the more interesting part of the question would be finding the complex solutions to $k$. While attempting this, I use assumptions:

First, the square root of a complex number with an imaginary part is always another complex number with an imaginary part. I don't see how this isn't true and I haven't found anything that says otherwise yet, so it seems good to me.

Second, if an equation with real and imaginary parts sums to $0$, then the real and imaginary parts alone will also sum to zero. I don't see how this is false either, since imaginary parts can't contribute to real parts, so both must just equal zero.

So in order to solve this problem, I set $k$ to equal $a+bi$ such that $a$ and $b$ are real numbers. Therefore,

$$\begin{align} &-2±\frac{\sqrt{3k^2-9k}}{k}\\ &=-2±\frac{\sqrt{3(a+bi)^2-9(a+bi)}}{a+bi}\\ &=-2±\frac{\sqrt{3(a^2+2abi-b^2)-9a-9bi)}}{a+bi}\\ &=-2±\frac{\sqrt{3a^2+6abi-3b^2-9a-9bi}}{a+bi}\\ \end{align}$$

Now if the numerator is equal to the denominator, then the fraction would equal 1 so I set it to that.

$$\begin{align} \sqrt{3a^2+6abi-3b^2-9a-9bi} &= a+bi\\ 3a^2+6abi-3b^2-9a-9bi &= (a+bi)^2\\ 3a^2+6abi-3b^2-9a-9bi &= a^2+2abi-b^2\\ 2a^2+4abi-2b^2-9a-9bi &= 0 \end{align}$$

Taking imaginary parts only:

$$\begin{align} 4abi-9bi &= 0\\ 4abi &= 9bi\\ 4ab &= 9b \end{align}$$

Assuming that $b ≠ 0$:

$$\begin{align} 4a &= 9\\ a &= \frac{9}{4} \end{align}$$

Now taking real parts only:

$$\begin{align}2a^2-2b^2-9a &= 0\\ 2\left(\frac{9}{4}\right)^2-2b^2-9\left(\frac{9}{4}\right) &= 0\\ 2\left(\frac{9}{4}\right)^2-9\left(\frac{9}{4}\right) &= 2b^2\\ \frac{486}{16} &= 2b^2\\ \frac{243}{16} &= b^2\\ ±\frac{9\sqrt{3}}{4} &= b \end{align}$$

Therefore $k = \frac{9}{4} + i\frac{9\sqrt{3}}{4}$ should be a solution. Whether or not cases like $k = \frac{9}{2} + i\frac{9\sqrt{3}}{2}$ that are multiples of the original answer would work don't really matter, but this seems like it should work. Another case that should also work would be setting the numerator of the original fraction to equal to zero. This also produces a complex solution for $k$.

The question is, can complex coefficients create real roots for quadratic equations, or did I calculate/assume/do the question wrong? If it is possible, did I miss any solutions? Any help would be appreciated.

  • $\begingroup$ i wrote the equation wrong nvm i go fix, it is fixed now $\endgroup$
    – John Liu
    Dec 12, 2020 at 0:30
  • $\begingroup$ "The question is, can complex coefficients create real roots for quadratic equations" Sure, why not? $(2+5i)x^3 - (7+5i)x^2+3x - 1$ will be $0$ if $x=1$. As all $x^k$ will be real, we must have the imaginary parts cancel out for $x=$ root but... no reason that can't happen. $\endgroup$
    – fleablood
    Dec 12, 2020 at 0:58
  • $\begingroup$ Oh, you said quadratic. No, I guess not trivially and not if all roots are real. If both roots, $r_1,r_2$ are real then $f(x) = a(x-r_1)(x-r_2)$ for some $a$. Unless $a$ is complex then all coefficients are real. (Of course if $a$ is complex that'd be a trivial example... but it's contrived.) $\endgroup$
    – fleablood
    Dec 12, 2020 at 1:12
  • $\begingroup$ "First, the square root of a complex number with an imaginary part is always another complex number with an imaginary part." true and worth explaining, with elementary complex analysis: for non-zero complex $z = re^{(i\theta)} = r[\cos(\theta) + i\sin(\theta)] : r \in \mathbb{R^+}, \theta \in (-\pi,\pi], ~\sqrt{z} = (\sqrt{r}) e^{(i\theta/2)}.$ The only non-zero complex $z$ with imaginary part $0$ will be $z$ of either of the forms $re^{(i\pi)} = (-r)$ or $re^{(i\times 0)} = r.$ $\endgroup$ Dec 12, 2020 at 1:47

2 Answers 2


You mention that we should be able to set the numerator of $\frac{\sqrt{3k^2 - 9k}}{k}$ to $0$. This would give $3k^2 = 9k$ and so $k = 3$, (discarding $k = 0$). This isn't complex so it isn't providing a solution to your question.

Other than that, your proposed solutions for $k$ do not seem to provide real solutions to the quadratic given.

Note that you can rewrite your quadratic as $k ( x^2 + 4x +1) = -9$. So we have $$k = \frac{-9}{ x^2 + 4x +1}$$ So if we plug in real numbers for $x$ (which would be the proposed solutions to your quadratic), then $k$ will also be real.

  • $\begingroup$ As others have mentioned, it is definitely possible to have real solutions to quadratic with complex coefficients, but for the specific case of polynomials of the form $kx^2+4kx+k+9=0$, there are no complex $k$ that give real solutions $x$ to the quadratic. $\endgroup$
    – wlpj2011
    Dec 12, 2020 at 16:11

To answer the more general question, "can complex coefficients lead to real-valued roots of polynomials," the answer is yes. We can readily see this by working backwards: construct a polynomial for which one root is real and the roots with nonzero imaginary part do not occur in complex conjugate pairs. Then the coefficients are guaranteed to not be pure real. For example, consider $$f(z) = (z - 3)(z - 2 - i) = z^2 - (5+i)z + (6+3i).$$ The coefficients are complex but by construction, there is a real root.

If we require all roots of a polynomial to be real-valued, then the answer is no unless we allow the polynomial to be a scalar multiple of some arbitrary complex constant. This follows from the fundamental theorem of algebra.


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