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I got the equation: $$68x^2 + 9 < 0,$$ and I did not know how to solve it. I tried getting it in standard quadratic form, but I could not factor it cleanly because there is no $x$ term. I'm so confused about how you even start to solve this. It is unlike anything I've seen before. Please help me. Thanks in advance!

P.S. I'm new to this forum

Blue
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Isaac Egan
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    Often when you have no idea how to even start a math problem a good advice is 'draw a picture'. If you graph the function and see when it is above and/or below zero that should give you some clue of what the answer is – Vincent May 13 '25 at 00:37
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    There is another tip I can give: when you feel that there is no $x$-term, you can imagine that there is an $x$-term, it just reads '0x'. However this advice is very specific to this type of problem, where my first advice 'draw a picture' is almost always a good idea, also in geometry, statistics etc – Vincent May 13 '25 at 00:38
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    The solutions to this are purely imaginary, and I am not super happy with this problem as such because the complex numbers do not have a nice order relationship. – Cameron L. Williams May 13 '25 at 01:04
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    I wonder if they expect you to say that there is no solution to this problem, but that would have to depend on your text and instructor. – Cameron L. Williams May 13 '25 at 01:07
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    It's subtle and I'd really worry about @CameronL.Williams observation that there is no order on the complex numbers BUT If $x = wi$ and either $w > \frac {3}{\sqrt{68}}$ OR $w < -\frac {3}{\sqrt{68}}$ then $w^2 > \frac 9{\sqrt{68}}$ and $x^w = (wi)^2 = w^2\cdot (-1) = -w^2 < -\frac 9{\sqrt{68}}$ is a perfectly valid solution. But for the LOVE OF THE MOTHER OF CLTHULU!!!!! *DON'T* make the mistake of trying to write it as $x > \frac 9{\sqrt{68}}i$!!!! You can NOT EVER put imaginary numbers in order! But you can put the real coefficients of the imaginary unit it order. – fleablood May 13 '25 at 01:19
  • This kind of problem makes perfect sense when $x$ is required to be real and the empty set (or "no solutions") is considered an acceptable answer when there are in fact no real solutions. If we saw this in context it might be clearer whether it is meant to be solved that way. – David K May 13 '25 at 01:43
  • The problem is it DOES have valid complex solutions. But in assigning it to a novice it encourages a very common and very WRONG and very DANGEROUS misconception. – fleablood May 13 '25 at 02:02
  • @fleablood If we solve it for complex number $x$, perhaps one may rephrase the problem as: Find all complex numbers $x$ such that $\mathrm{Im}(68x^2+9) = 0$ and $\mathrm{Re}(68x^2 + 9) < 0$. If we solve it for real number $x$, 'no solution'. – River Li May 13 '25 at 05:15
  • @RiverLi But... if you ask the question quote unquote "Find all numbers (complex or real) so that $68x^2 + 9 < 0$" that question does have a solution. The first thing to notice is $68x^2 + 9 < 0$ can only be true if $68x^2 + 9$ is a real number and then... It may be a weird question and one no-one would ever ask but it is a valid question. FWIW the sentence "$(2 + 3i) + (4-i) - (1+2i) > 3$" is a true sentence even though it it should be taken to a wood shed and beaten. – fleablood May 13 '25 at 06:56
  • @fleablood In my rephrasing, all expressions before $< 0$ or $> 0$ are real-valued expression for all complex numbers, i.e. $\mathrm{Im}(68x^2 + 9)$ or $\mathrm{Re}(68x^2 + 9)$. However, the expression $68x^2 + 9$ are not real-valued expression for all complex numbers $x$. – River Li May 13 '25 at 10:12

3 Answers3

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$68x^2 + 9 < 0\iff$

$68x^2 < -9\iff$

$x^2 < -\frac {9}{68}$

So $x^2 < -\frac {9}{68} < 0$ then $x^2$ must be a square number that is negative.

Now if we are only using real numbers and not complex number then.... No number when squared is negative. So there is no solution. This is simply impossible.

[I don't know what text or course this is for but it is probably intended we are working with real number only. I'm not sure if I agree with Thomas Andrews that it is "almost certain" we are working in real numbers only, but it is certainly the more probable.]

.....

[But just in case we are including Complex numbers (which we probably arent) I find it interesting that there is a distinct set of solutions. So... the complex number solutions: ]

If we are using complex numbers then....

we have observations. $i^2 = -1$ and $(-i)^2 = (-1)^2i^2 = -1$. And if we have $z = a + bi$ where $a, b$ are real then $(a+bi)^2 = (a^2-b^2) + 2ab i$ where $a^2 -b^2$ and $2ab$ are real.

If $b = 0$ then $z$ is a purely real number and $z=a$ and $z^2 = a^2 \ge 0$. Positive is $a\ne 0$ and never negative.

If $a=0$ then $z = bi$ is purely imaginary and $z^2 = -b^2 \le 0$. Negative if $b\ne 0$ and never positive.

If neither $a$ nor $b= 0$ then $z^2 = (a^2 -b^2) + 2ab i$ not a real number. It is a complex number and complex numbers are neither positive nor negative. (Indeed it is impossible to order complex numbers in such a way that one complex number is "$<$" or "$>$" than another.)

So if we have $x^2 < 0$ the $x = bi$, a purely imaginary number.

If $x = bi$ where $b$ is real then $x^2 = (bi)^2 = b^2(-i)^2 = -b^2$.

So we have $x^2 = -b^2 < -\frac 9{68} < 0$ and $b^2 > \frac 9{68} > 0$. Now we can solve for $b$ just like any other real number.

$b^2 > \frac 9{68} \implies b > \frac 3{\sqrt{68}}$ OR $b < -\frac 3{\sqrt {68}}$.

So the solution is $x$ is a purely imaginary number $x=bi$ where the real value of the imaginary component $b$ is such that $b > \frac 3{\sqrt{68}}$ or $b < -\frac 3{\sqrt{68}}$.

Now it's important to realize: IT IS IMPOSSIBLE TO COMPARE THE ORDER OF COMPLEX NUMBERS AND SO ONE IS "LARGER" THAN ANOTHER. Even with purely imaginary ones we can not do that. We can make statementa about the comparative size of the components of complex numbers as the components themselves are real, but not the complex numbers themselves.

fleablood
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  • You should make clear that the case of $x$ real is almost certainly intended. I have never seen an inequality for a complex number formula, except for ones that only involve $|\cdot|$ or the real and imaginary part functions to get real numbers. Your answer feels like it is treating the two as equally likely. – Thomas Andrews May 13 '25 at 03:34
  • Okay. If you read it as such. I'd argue a question as stated should be answered as stated and ... well, it's moot. – fleablood May 13 '25 at 04:15
  • I took "imaginary solutions" to mean the solutions to the corresponding equation $=0,$ which is the usual way to start quadratic inequality problems. They never asked to solve for complex $x.$ (You have to check the history to see the actual reference, in the title: "How to solve a quadratic inequality with imaginary solutions? (68x^2 + 9 < 0)" – Thomas Andrews May 13 '25 at 05:20
  • If we can't order complex numbers, then how can we find the solution of this complex quadratic inequality? Also, could I just take the square root of both sides when the equation is x^2 < -9/68? Then, it would become +/- x = +/- 3i/sqrt(68). Then, you could test values to see which categories actually work. Would this way of doing the problem actually work? – Isaac Egan May 13 '25 at 23:13
  • We can't order complex numbers but we can order real numbers. And we can describe complex numbers in terms of properties of their components. In this case $x^2$ must be a real negative number (notice that is a statement about $x$ squared; not about $x$). In order for a number squared to be negative the number itself must be purely imaginary. So $x = 0 + wi$ for some real $w$ while $x$ is not real but purely imaginary. We have $(wi)^2 = -w^2 < -\frac{9}{68}$. That is an inequality about REAL numbers and has nothing to do with complex numbers... to be cont.... – fleablood May 14 '25 at 01:35
  • So as $-w^2 < -\frac 9{68}$ then $w^2 > \frac 9{68}$ so $w$ which remember $w$ is a REAL number is such that $w > \sqrt{\frac 9{68}}$ or $w < -\sqrt{\frac 9{68}}$. Notice we NEVER did any inequality involving complex numbers... "Also, could I just take the square root of both sides when the equation" No. $x^2 < a$ does NOT mean $x< \pm \sqrt a$ and that just doesn't make any sense. – fleablood May 14 '25 at 01:39
  • But as Thomas Andrews pointed out. It is almost certain that the intended solution is "There is not solution". Putting something in terms of $f(x) < k$ means that $f(x)$ and $k$ MUST both be real numbers and this very strongly implies the input of $x$ is assumed to be real (it is very rare to refer to $f(x)$ where $x$ may be complex but was are restricting it only to real output without explicitly stating that is what we are doing) in which case the solution to $x^2 < 0$ when $x$ is real is "THERE IS NO SOLUTION". Period. – fleablood May 14 '25 at 01:49
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You can first just take a look at the equality version of this equation and use the quadratic formula to solve this, and then test the (usually three) regions this brings up. Solving for $68x^2 + 9 = 0$, we have $a = 68, b = 0, c = 9$ and so the quadratic equation is just $x = \frac{\pm\sqrt{-2448}}{136}$ which means $x$ is $\frac{\pm\sqrt{2448}}{136}i$, which means the inequality has only complex solutions. When this is the case you can't really represent the answer as an inequality because inequalities are only defined for numbers in the real plane. Another math stack exchange question covers why this is the case better than I can (changed the example due to a comment identifying a better one): Total ordering on complex numbers

Rishiit Sharma
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  • Although $x$ is purely imaginary, you can express $x$ as $x = wi$ where $w$ (NOT $x$) is real and you can express $x$ as $x = wi: w> \frac {\sqrt{2448}}{136}$ OR $w < \frac {\sqrt{2448}}{136}$. But you can not say "$x > \frac {\sqrt{2448}}{136}i$" or "$x <- \frac {\sqrt{2448}}{136}i$" (or "$x < \frac {\sqrt{2448}}{136}i$" or "$x >- \frac {\sqrt{2448}}{136}i$") – fleablood May 13 '25 at 02:00
  • IMO the cited page fails to explain why inequalities can exist for complex numbers (as alegbraically consistant). A better (but not perfect) post would be https://math.stackexchange.com/questions/487997/total-ordering-on-complex-numbers – fleablood May 13 '25 at 02:27
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The inequality $68x^2 + 9 < 0$ is likely intended for real numbers. Since $68x^2 \geq 0$, we have

$68x^2 + 9 \geq 9 > 0$

To confirm, solve the equality $68x^2 + 9 = 0$:

$68x^2 = -9 \implies x^2 = -\frac{9}{68}$

Using the quadratic formula (with $a = 68$, $b = 0$, $c = 9$):

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{0 \pm \sqrt{0 - 4 \cdot 68 \cdot 9}}{2 \cdot 68} = \frac{\pm \sqrt{-2448}}{136} = \pm \frac{\sqrt{2448}}{136} i = \pm \frac{3\sqrt{17}}{34} i$

The roots are complex, so no real roots exist, and testing regions on the real number line isn’t applicable since

$68x^2 + 9 > 0 \quad \text{everywhere}$

For complex solutions (as you asked), consider $x = bi$ (purely imaginary):

$x^2 = -b^2 < -\frac{9}{68} \implies b^2 > \frac{9}{68} \implies |b| > \sqrt{\frac{9}{68}} \approx 0.364$

Thus, complex solutions are $x = bi$ with $|b| > \sqrt{\frac{9}{68}}$.

Note that inequalities for complex numbers are non-standard and typically involve $|\cdot|$ or real/imaginary parts, so this interpretation assumes $x^2$ is real and negative.

Amit Mittal
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  • You should make clear that the case of $x$ real is almost certainly intended. I have never seen an inequality for a complex number formula, except for ones that only involve $|\cdot|$ or the real and imaginary part functions to get real numbers. Your answer feels like it is treating the complex case is most likely. – Thomas Andrews May 13 '25 at 03:36
  • @ThomasAndrewsYou’re right that inequalities with complex numbers are uncommon outside of contexts involving $(|\cdot|)$ or real/imaginary parts, and the real case is likely intended. Since $(68x^2 + 9 > 0)$ for all real $(x)$, the inequality has no real solutions. I explored the complex case because the user specifically asked for it, but I’ll clarify that the real case is typically the focus for such problems unless otherwise specified. Thanks for pointing this out! – Amit Mittal May 13 '25 at 04:46
  • @ThomasAndrews I edit my answer – Amit Mittal May 13 '25 at 05:04