I provide an alternative trigonometric formula for solving quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.

Theorem 1. If $ax^2+bx+c=0$, then


where $\theta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Multiplying by $a$ the quadratic equation we have


Let's make the substitution $\sec{\theta}=\frac{b}{2\sqrt{ac}}$, then

$$\begin{aligned}0&=(ax)^2+b(ax)+ac\\&=(ax)^2+2\sqrt{ac}(ax)\sec{\theta}+ac\\&=\sec{\theta}\left((ax)^2\cos{\theta}+2\sqrt{ac}(ax)+ac\cos{\theta}\right)\\&=\cos{\theta}((ax)^2+ac)+2\sqrt{ac}(ax)\\&= \cos{\theta}(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\cos{\theta}+2\sqrt{ac}(ax)\\&= \left(\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\right)(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\left(1-2\sin^2{\frac{\theta}{2}}\right)+2\sqrt{ac}(ax)\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 +4\sqrt{ac}(ax)\sin^2{\frac{\theta}{2}}\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}\left((ax+\sqrt{ac})^2 -4\sqrt{ac}(ax)\right)\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax-\sqrt{ac})^2\\&= \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) +\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right) \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) -\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right)\\&=\left(ax\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right) \right) \left(ax\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right) \right).\end{aligned}$$

Now, by setting the factors equal to zero and solving for x, we obtain,

$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}}\right)\sqrt{\frac{c}{a}}\\&= \left(1+\frac{2}{\tan{\frac{\theta}{2}-1}}\right)\sqrt{\frac{c}{a}}. \end{aligned}$$

Similarly we obtain,

$$\begin{aligned}x_2= \left(1-\frac{2}{\tan{\frac{\theta}{2}+1}}\right)\sqrt{\frac{c}{a}} \end{aligned}.$$

The formula appears to be new, at least on the internet. Wikipedia presents a trigonometric method, but it is different from mine. Stuart Simons offers a method that does seem to be related to my approach, but he only provides direct formulas for complex roots. I was able to independently derive Simons' formulas using $\cos{\theta}$ in the substitution instead of $\sec{\theta}$, and this was before coming across the Wikipedia article that references Simons' paper: Simons, Stuart, 'Alternative approach to complex roots of real quadratic equations,' Mathematical Gazette 93, March 2009, 91–92.

EDITED. And this formula can be obtained by substituting $\sin{\theta}$ instead of $\sec{\theta}$ and following similar steps as my previous proof: $x_{1,2}=\pm i e^{\pm i \theta}\sqrt{\frac{c}{a}}$.

Note. My interest in these formulas is purely mathematical. My primary intention is to incorporate all of these formulas mentioned here into a unified framework based on half-angle formulas. ALL the formulas mentioned here by me and others can be derived using half-angle identities, as exemplified by my initial proof, so in that sense, I feel satisfied.

The key observation I made is that $a^2+2abf(x)+b^2$ can be nicely factorized thanks to the properties of half-angle formulas, as long as $f(x)$ is a trigonometric function except $\tan{\theta}$ or $\cot{\theta}$. I wouldn't be surprised if analogous formulas could be obtained for cubic and quartic equations, but at the moment, I have no idea how to do it.

EDITED. Here I provide a proof for the formulas cited by njuffa in the comments.

Theorem 2.Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then

$$\begin{aligned}x_{1}&=-\tan{\frac{\theta}{2}}\sqrt{\frac{c}{a}},\\ x_{2}&=-\cot{\frac{\theta}{2}}\sqrt{\frac{c}{a}}, \end{aligned}$$

where $\theta=\csc^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Multiplying by $a$ the quadratic equation $ax^2+bx+c=0$, we have


Let's make the substitution $\csc{\theta}=\frac{b}{2\sqrt{ac}}$, then

$$\begin{aligned}0&=(ax)^2+b(ax)+ac\\&=(ax)^2+2\sqrt{ac}(ax)\csc{\theta}+ac\\&=\csc{\theta}\left((ax)^2\sin{\theta}+2\sqrt{ac}(ax)+ac\sin{\theta}\right)\\&=2(ax)^2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}+2\sqrt{ac}(ax)\left(\sin^2{\frac{\theta}{2}}+\cos^2{\frac{\theta}{2}}\right)+2ac\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}\\&=(ax)^2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}+\sqrt{ac}(ax)\sin^2{\frac{\theta}{2}}+ \sqrt{ac}(ax)\cos^2{\frac{\theta}{2}}+ac\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}\\&= ax\sin{\frac{\theta}{2}}\left(ax\cos{\frac{\theta}{2}}+\sqrt{ac}\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\cos{\frac{\theta}{2}}\left(ax\cos{\frac{\theta}{2}}+\sqrt{ac}\sin{\frac{\theta}{2}}\right)\\&= \left(ax\cos{\frac{\theta}{2}}+\sqrt{ac}\sin{\frac{\theta}{2}}\right)\left(ax\sin{\frac{\theta}{2}}+\sqrt{ac}\cos{\frac{\theta}{2}}\right)\end{aligned}$$ Finally, by setting the factors equal to zero, we obtain the desired formulas.

The following Euler-like identities have been suggested by formulas in this question.

If complex $\theta_1=\cos^{-1}{(p)}$ and $\theta_2=\sec^{-1}{(p)}$, where $p\geq-1$ and $p\neq0$, then the following relation holds:

$$e^{i\theta_1}=\frac{1-\tan{\frac{\theta_2}{2}} }{1+\tan{\frac{\theta_2}{2}}} \tag{1}$$

And when $p<-1$, we have


I came up with $(1,2)$ by equating the formulas for the positive roots of the original question's formula with Simons' formulas. Then, they need a slight modification because we cancel out $\sqrt{\frac{c}{a}}$. Another simpler formula can be obtained by equating the formulas for the roots of the german text and the formulas I got from substituting by $\sin{\theta}$ instead of $\sec{\theta}$:

If complex $\theta_1=\sin^{-1}{(p)}$ and $\theta_2=\csc^{-1}{(p)}$, where $p\leq1$ and $p\neq0$, then the following relation holds:


And for $p>1$ we have


There are several variants we can get this way. I find identities $(1-4)$ interesting because they allow for the simplification of certain trigonometric integrals, as well as solving integrals that not even Mathematica can handle. For example,

$$\int_{2}^{3} \frac{{1 - \tan\frac{{\sec^{-1}x}}{2}}}{{1 + \tan\frac{{\sec^{-1}x}}{2}}}\sqrt{\tan\frac{\csc^{-1}x}{2}}\,dx$$

Question: Is this formula known? Did Simons consider it in his article (I don't have access to it)?

  • 8
    $\begingroup$ This formula seems way more complicated than the traditional quadratic formula. It seems also very impractical to apply even in some very simple cases like $x^2-1=0$. Do you have any reason to want such a obfuscated formula? $\endgroup$
    – jjagmath
    Jan 13 at 20:31
  • 2
    $\begingroup$ @EmmanuelJoséGarcía: The Simons article is available through JSTOR, which allows anyone to read 100 articles per month for free. (Select "Alternate Access Options".) Some colleges and universities offer access as well, even to alumni. $\endgroup$
    – Blue
    Jan 13 at 23:07
  • 1
    $\begingroup$ I would suggest your proof could either be shorter or more instructive: you could shorten it greatly by simply showing $x_1+x_2=-\frac{b}{a}$ and $x_1 x_2=\frac{c}{a}$ for the formulae you gave, or you could motivate the choice of $\theta$. A geometric proof might be nice, but I don't see an obvious one. Regarding the motivation (which is arguably more interesting), it might be easier to first consider monic quadratics (since it's easy to convert to the general case). $\endgroup$ Jan 13 at 23:19
  • 2
    $\begingroup$ @EmmanuelJoséGarcía In the 19th century, when all computation was manual (assisted by tables), solving quadratic equations by a trigonometric method was recommended when the coefficients were large, comprising many digits. The details differ from what is shown here, but usually $x_{1}$ was compute via $\tan \frac{\alpha}{2}$ and $x_{2}$ via $\cot \frac{\alpha}{2}$. For an example see here (in German) $\endgroup$
    – njuffa
    Jan 14 at 10:23
  • 1
    $\begingroup$ @EmmanuelJoséGarcía An example of the trigonometric solution of a quadratic equation in an English-language publication $\endgroup$
    – njuffa
    Jan 14 at 10:37

4 Answers 4


If the formulas are to be understood in the context of real trigonometry (rather than complex trigonometry), we must assume that $a$ and $c$ have the same sign; otherwise we would not be able to take the square root of $\dfrac ca.$ So in this answer I will consider only this case.

Let $\alpha = \dfrac{b}{2a}$ and $\beta = \sqrt{\dfrac ca}.$ Then $\dfrac{b}{2\sqrt{ac}} = \dfrac\alpha\beta$ and $ax^2 + bx + c$ is equivalent to $$ x^2 + 2\alpha x + \beta^2. $$

The roots of the quadratic, $x_1$ and $x_2,$ both have the opposite sign from $\alpha.$ Their sum is $-2\alpha$ and their product is $\beta^2.$ So we can assign the magnitudes of these various quantities to parts of the semicircle shown below:

enter image description here

That is, the radius of the semicircle is $\lvert\alpha\rvert,$ the diameter is $\lvert x_1\rvert + \lvert x_2\rvert = \lvert2\alpha\rvert,$ and if we place point $C$ on the diameter so that $AC = \lvert x_1\rvert$ and $BC = \lvert x_2\rvert$ and construct $CD$ perpendicular to $AB$ with $D$ on the semicircle, $CD = \beta = \sqrt{x_1 x_2}.$

Consider the case where $\alpha > 0.$ If we then let $\theta = \angle CDO,$ then $\theta = \operatorname{arcsec}\left(\dfrac\alpha\beta\right) = \operatorname{arcsec}\left(\dfrac{b}{2\sqrt{ac}}\right).$ Then $\angle COD = \dfrac\pi2 - \theta$ and by the inscribed angle theorem, $\angle BDC = \angle CAD = \dfrac\pi4 - \dfrac\theta2.$ Therefore $\lvert x_1\rvert = \beta \cot\left(\dfrac\pi4 - \dfrac\theta2\right)$ and $\lvert x_2\rvert = \beta \tan\left(\dfrac\pi4 - \dfrac\theta2\right),$ but since $x_1$ and $x_2$ are both negative in this case, $$ x_1 = -\beta \cot\left(\dfrac\pi4 - \dfrac\theta2\right) \quad\text{and}\quad x_2 = -\beta \tan\left(\dfrac\pi4 - \dfrac\theta2\right). $$

According to the formula for the tangent of a difference of angles, $$ \tan\left(\dfrac\pi4 - \dfrac\theta2\right) = \frac{\tan\dfrac\pi4 - \tan\dfrac\theta2}{1 + \tan\dfrac\pi4 \tan\dfrac\theta2} = \frac{1 - \tan\dfrac\theta2}{1 + \tan\dfrac\theta2}. $$

The formulas for the roots then become $$ x_1 = \frac{\tan\dfrac\theta2 + 1}{\tan\dfrac\theta2 - 1} \sqrt{\dfrac ca} \quad\text{and}\quad x_2 = \frac{\tan\dfrac\theta2 - 1}{\tan\dfrac\theta2 + 1} \sqrt{\dfrac ca}, $$ which can be manipulated algebraically into to the formulas shown in the question.

In the case where $\alpha < 0,$ then we can reassign the angles in the figure so that $\angle CDO = \pi - \theta$ where $\theta = \operatorname{arcsec}\left(\dfrac{b}{2\sqrt{ac}}\right).$ Then $\angle COD = \theta - \dfrac\pi2$ and $\angle CAD = \dfrac\theta2 - \dfrac\pi4$. But $$ \tan\left(\dfrac\theta2 - \dfrac\pi4\right) = -\tan\left(\dfrac\pi4 - \dfrac\theta2\right) = \frac{\tan\dfrac\theta2 - 1}{\tan\dfrac\theta2 + 1}, $$ and since $x_1$ and $x_2$ are positive this time, after some algebra we get the same formulas for them in the end as before.

The trigonometric formula in the Wikipedia article for the "positive" case is based on the same figure, but with $\angle COD = 2\theta_p.$ Your formula is therefore just a variation that presumably could be obtained from the formula in Wikipedia by trigonometric identities, although it's not an obvious derivation. Also note that your formula requires more calculation, which is one reason you are less likely to find it documented as a solution of a quadratic equation (given the traditional use of trigonometric substitution in table-assisted calculation).

Another difference between your approach and the one in Wikipedia is that the Wikipedia formula requires using two solutions of the equation $\sin(2\theta) = -2 \dfrac{\sqrt{ac}}{b},$ whereas you use only the pseudoinverse function arcsec, which provides only one value, and then you use sign changes to get the two roots. So your nethod is more explicit, although I think I would prefer it in the form $$ \frac{\tan\dfrac\theta2 \pm 1}{\tan\dfrac\theta2 \mp 1} \sqrt{\frac ca} $$ because it is a bit more obvious how the trigonometric formula in the "$+$" case is the reciprocal of the "$-$" case.

To handle the case where the first and last coefficients of the quadratic have opposite signs, you will either need to modify your formulas or go to trigonometric functions of complex numbers, which makes the geometric derivation more difficult but can still be treated algebraically.

As a side note, in the figure above, $OC = \sqrt{\alpha^2 - \beta^2}.$ If $a > 0$ and $b < 0,$ we then get $$ x_{1,2} = \lvert\alpha\rvert \pm \sqrt{\alpha^2 - \beta^2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$ the well-known solution of the quadratic equation.

  • $\begingroup$ Thank you for providing an alternative derivation. I deemed it unnecessary to assume that $a$ and $b$ have the same sign, as not assuming so would lead to a product of complex numbers that must yield the roots. For example, consider the equation $x^2-2x-15=0$. Wolfram Alpha gives one of the roots using my formula. $\endgroup$ Jan 15 at 11:59
  • $\begingroup$ Of course, this is one of those cases where I would not use my formula and would prefer to use the traditional formula or factorization. $\endgroup$ Jan 15 at 12:00
  • $\begingroup$ When you say that my formula requires more calculation, do you mean that deriving it is more complicated than the formulas in the Wikipedia article, or that my formula involves more steps when calculating roots compared to the formulas in the article? $\endgroup$ Jan 15 at 12:02
  • $\begingroup$ Either way, if it's the latter, my purpose in deriving this formula is not to streamline calculations like 19th-century astronomers but to incorporate them into a unified theoretical framework based on half-angle formulas. As mentioned earlier, by using substitutions such as $\cos{\theta}$ and $\csc{\theta}$ instead of $\sec{\theta}$, you can obtain the Simons formulas mentioned above, as well as the formulas cited by njuffa, respectively. $\endgroup$ Jan 15 at 12:02

In this answer I would like to compare the proposed solution with the method documented in the section of the Wikipedia article discussing trigonometric solutions of a quadratic equation.

According to Wikipedia, in order to solve $ax^2 + bx + c$ with the signs of $a$ and $c$ the same, we first solve for $\theta_p$ in $\sin(2 \theta_p) = -2 \frac{\sqrt{ac}}{b}.$ There actually is a half-angle formula hidden in this method, as we can see by making the substitution $\phi = 2\theta_p.$ The equation to solve is then $$ \sin\phi = -2 \frac{\sqrt{ac}}{b}. \tag1 $$ We then plug each of the solutions of this equation into the formula $$ x = \sqrt{\frac ca} \tan \frac\phi2. $$ And there is a simple half-angle formula to solve a quadratic equation.

The two roots $x_1$ and $x_2$ arise from distinct solutions of Equation $(1).$ The solutions of this equation are all in one or the other of the the following two forms, \begin{align} \phi_1 &= \arcsin\left(-2 \frac{\sqrt{ac}}{b}\right) + 2k\pi, & \phi_2 &= \pi - \phi_1, \\ \end{align} where $k$ is any integer. But then any two possible values of $\frac{\phi_1}2$ differ by a multiple of $\pi,$ and $\tan$ is a periodic function with period $\pi,$ so the value of $\tan \frac{\phi_1}2$ is independent of the choice of $k.$ For the same reasons, the value of $\tan \frac{\phi_2}2$ also is independent of the choice of $k.$ So we can take $k = 0$ for two representative solutions, \begin{align} x_1 &= \sqrt{\frac ca} \tan \frac\phi2, & x_2 &= \sqrt{\frac ca} \tan \frac{\pi-\phi}2 \end{align} where $$ \phi = \arcsin\left(-2 \frac{\sqrt{ac}}{b}\right) = -\arcsin\left(2 \frac{\sqrt{ac}}{b}\right). $$

Alternatively, noting that $\tan\left(\frac\pi2 - \alpha\right) = \cot\alpha,$ we could write \begin{align} x_1 &= \sqrt{\frac ca} \tan \frac\phi2, & x_2 &= \sqrt{\frac ca} \cot \frac\phi2. \end{align}

This seems like a very nice use of half-angle trigonometry to me. Note that this solution is also found explicitly using the tangent of a half angle in 19th-century texts such as this (where $\alpha$ is the same as my $\phi$) and this (whose $\alpha$ is equal to $-\phi$ in my solution).

Since $\arcsin(x) = \frac\pi2 - \arccos(x),$ yet another way to write $\phi$ in the above solution is $$ \phi = \arccos\left(2 \frac{\sqrt{ac}}{b}\right) - \frac\pi2. $$ So if we let $$ \theta = \operatorname{arcsec}\left(\dfrac{b}{2\sqrt{ac}}\right) = \arccos\left(2 \frac{\sqrt{ac}}{b}\right) $$ the solutions are \begin{align} x_1 &= \sqrt{\frac ca} \tan\left(\dfrac\theta2 - \dfrac\pi4\right), & x_2 &= \sqrt{\frac ca} \cot\left(\dfrac\theta2 - \dfrac\pi4\right), \end{align} which as we can see in my answer posted earlier are equivalent to the "$+$" and "$-$" cases of the formula in the question, although subtracting $\frac\pi4$ from each half-angle in the formulas leads to a relatively complicated expression in terms of $\tan\frac\theta2.$


As suggested in a comment, here's how to prove your formula using Vieta's formula. In the quadratic equation $ax^2 + bx + c = 0,$ the two roots $x_1,x_2$ satisfy the two equations $$ x_1 + x_2 = -\frac ba, \\ x_1 x_2 = \frac ca. $$ But conversely, if we find $x_1,x_2$ that satisfy the two equations above then we have found the roots of $ax^2 + bx + c = 0.$

Let $ax^2 + bx + c = 0$ and let $$ x_{1,2} = \left(1\pm\frac2{\tan\frac{\theta}2 \mp 1}\right)\sqrt{\frac ca} \tag1 $$ where $\theta = \operatorname{arcsec}\left(\dfrac{b}{2\sqrt{ac}}\right).$ We wish to show that $x_1,x_2$ are the roots of $ax^2 + bx + c = 0.$

It will be convenient to rewrite the bracketed factor in Equation $(1)$ as follows: $$ 1\pm\frac{2}{\tan{\frac{\theta}2 \mp 1}} = \frac{\left(\tan\frac{\theta}2 \mp 1\right) \pm 2}{\tan\frac{\theta}2 \mp 1} = \frac{\tan\frac{\theta}2 \pm 1}{\tan\frac{\theta}2 \mp 1}. $$

Then \begin{align} x_1 + x_2 &= \frac{\tan\frac{\theta}2 + 1}{\tan\frac{\theta}2 - 1}\sqrt{\frac ca} + \frac{\tan\frac{\theta}2 - 1}{\tan\frac{\theta}2 + 1}\sqrt{\frac ca} \\ &= \frac{(\tan\frac{\theta}2 + 1)^2 + (\tan\frac{\theta}2 - 1)^2} {\tan^2\frac{\theta}2 - 1}\sqrt{\frac ca} \\ &= \frac{2(\tan^2\frac{\theta}2 + 1)}{\tan^2\frac{\theta}2 - 1}\sqrt{\frac ca}. \end{align}

Now, using the identity $$ \tan^2\frac{\theta}2 = \frac{1 - \cos\theta}{1 + \cos\theta}, $$ we have \begin{align} \tan^2\frac{\theta}2 + 1 &= \frac{(1 - \cos\theta) + (1 + \cos\theta)}{1 + \cos\theta} = \frac{2}{1 + \cos\theta}, \\ \tan^2\frac{\theta}2 - 1 &= \frac{(1 - \cos\theta) - (1 + \cos\theta)}{1 + \cos\theta} = \frac{-2\cos\theta}{1 + \cos\theta}, \end{align}

and therefore \begin{align} x_1 + x_2 &= \frac{2\left(\frac{2}{1 + \cos\theta}\right)} {\left(\frac{-2\cos\theta}{1 + \cos\theta}\right)} \sqrt{\frac ca} \\ &= -2 \sec\theta \sqrt{\frac ca} \\ &= -2 \left(\dfrac{b}{2\sqrt{ac}}\right) \sqrt{\frac ca} \\ &= -\frac ba. \tag2 \end{align}

Also, $$ x_1 x_2 = \left(\frac{\tan\frac{\theta}2 + 1}{\tan\frac{\theta}2 - 1}\sqrt{\frac ca}\right) \left(\frac{\tan\frac{\theta}2 - 1}{\tan\frac{\theta}2 + 1}\sqrt{\frac ca}\right) = \frac ca. \tag3 $$

From Equations $(2)$ and $(3)$ it follows that $x_1,x_2$ are the roots of $ax^2 + bx + c = 0.$

  • $\begingroup$ In that case, it wouldn't be a "derivation" since you have to know the roots $x_1$ and $x_2$ beforehand. Judging by your proof, I don't think the "simplification" is significant enough to justify discarding a true derivation that only relies on the half-angle formulas, completing the square, and factoring the difference of squares. $\endgroup$ Jan 20 at 16:06
  • $\begingroup$ Additionally, my approach is more holistic in the sense that by changing the substitution to any of the other trigonometric functions (except tangent or cotangent), you obtain either the formulas mentioned by njuffa, Simons' formulas, or the one I have recently added to my original question. A more panoramic view is achieved through my approach; therefore, I prefer my proof over the others. $\endgroup$ Jan 20 at 16:06
  • 1
    $\begingroup$ It's true that this is not a derivation. As I said, it was motivated by a comment. It is a confirmation of the derivation, much as one could confirm a lengthy solution of an integral by differentiating the result. I omitted the derivation since you have already shown it, but I do not suggest that you should have omitted the derivation from the question, since that is (after all) the motivation of your solution. A reason one might add a proof like this to the exposition of the formula is that it can be less work for someone to verify the proof than to verify the derivation. $\endgroup$
    – David K
    Jan 20 at 20:09

I found the derivation somewhat difficult to follow due to the length of the formulas. MathJax also seems to have some difficulty formatting it correctly in my browser, possibly also because of the length. So I wondered, is it possible to do the derivation more succinctly?

For given parameters $a,b,c$ with $a \neq 0,$ we want to find the two (possibly complex, possibly repeated) values of $x$ that solve $ax^2+bx+c=0$.

Multiply by $a$ to obtain $a^2 x^2 + abx + ac$ and make the substitutions $y = ax$ and $r^2 = ac$ (using a suitable complex square root of $ac$ if necessary): $$ y^2 + by + r^2 = 0. $$

Now substitute $b = 2r \sec\theta.$ Use complex $\theta$ if necessary, because real $\theta$ is suitable only when $b = 2ru$ for real $u$ and $\lvert u\rvert \geq 1.$ (Note that this substitution requires that either $b = c = 0$ or that $b$ and $c$ are both non-zero, since even with complex $\theta$ we have $\sec\theta \neq 0.$ We cannot use this substitution to solve $ax^2 + c = 0$ or $ax^2 + bx = 0.$) Then $$ y^2 + 2ry\sec\theta + r^2 = 0. $$

Perform a Weierstrass substitution (which has a long history, related even to classical Greek mathematics), $t = \tan\dfrac\theta2,$ $\cos\theta = \dfrac{1-t^2}{1+t^2},$ hence $\sec\theta = \dfrac{1+t^2}{1-t^2}.$

$$ y^2 + 2ry \frac{1+t^2}{1-t^2} + r^2 = 0. $$

Multiply by $1 - t^2.$

$$ \left(y^2 + r^2\right)\left(1 - t^2\right) + 2ry \left(1 + t^2\right) = 0. $$

Collect terms in $t^2.$

$$ \left(y^2 + 2ry + r^2\right) - \left(y^2 - 2ry + r^2\right) t^2 = 0. $$


$$ (y + r)^2 - (y - r)^2 t^2 = 0. $$

Factorize again.

$$ \left((y + r) + (y - r) t\right)\left((y + r) - (y - r) t\right) = 0. $$

Within each factor, collect terms in $y.$

$$ \left((1 + t)y + (1 - t)r\right)\left((1 - t)y + (1 + t)r\right) = 0. $$

The two solutions in $y$ are the values of $y$ that make respectively the first or second factor zero: $$ y_1 = \frac{1 - t}{1 + t} r, \quad y_2 = \frac{1 + t}{1 - t} r. $$

To find the two solutions in $x,$ reverse the substitution $y = ax,$ divide each side of each equation by $a,$ and use the fact that $\left(\dfrac ra\right)^2 = \dfrac ca$ to write $\dfrac ra = \sqrt{\dfrac ca}$: $$ x_1 = \frac{1 - t}{1 + t} \sqrt{\frac ca}, \quad x_2 = \frac{1 + t}{1 - t} \sqrt{\frac ca}. $$

Reverse the substitution $t = \tan\dfrac\theta2.$ $$ x_1 = \frac{1 - \tan\frac\theta2}{1 + \tan\frac\theta2} \sqrt{\frac ca}, \quad x_2 = \frac{1 + \tan\frac\theta2}{1 - \tan\frac\theta2} \sqrt{\frac ca}. $$

These expressions can then easily be rearranged into the forms shown in the original question.

You could, in fact, use $\tan\dfrac\theta2$ instead of $t$ throughout the derivation, using the trigonometric identity $$\sec\theta = \dfrac{1+\tan^2\frac\theta2}{1-\tan^2\frac\theta2},$$ but I like using substitution to keep formulas concise during extensive algebraic manipulations.

Many of the steps in this derivation correspond with steps in the derivation in the original question. The main difference lies in immediately replacing $\sec\theta$ via a half-angle identity (the Weierstrass substitution), which ends up simplifying the other steps a bit. Also, as a matter of style, I prefer to avoid writing equations such as $$\sec{\theta}\left((ax)^2\cos{\theta}+2\sqrt{ac}(ax)+ac\cos{\theta}\right)=\cos{\theta}((ax)^2+ac)+2\sqrt{ac}(ax)$$ even when they happen to be true (by virtue of the left side already being known equal to zero), but most of the steps of this derivation (like the steps in the original derivation) could be written as a long string of equations if desired.


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