Mathematician Finds Easier Way to Solve Quadratic Equations

[Thomas Klemm]

Double Angle Formulae

We start with the double angle formula for the sine:

\(
\begin{align}
\sin(2 \alpha)
&= \frac{2 \sin(\alpha) \cos(\alpha)}{1} \\
\\
&= \frac{2 \sin(\alpha) \cos(\alpha)}{\cos(\alpha)^2 + \sin(\alpha)^2} \\
\\
&= \frac{2 \tan(\alpha)}{1 + \tan(\alpha)^2} \\
\end{align}
\)

And similarly for the tangent:

\(
\begin{align}
\tan(2 \alpha)
&= \frac{\sin(2 \alpha)}{\cos(2 \alpha)} \\
\\
&= \frac{2 \sin(\alpha) \cos(\alpha)}{\cos(\alpha)^2 - \sin(\alpha)^2} \\
\\
&= \frac{2 \tan(\alpha)}{1 - \tan(\alpha)^2} \\
\end{align}
\)

Therefore both can be calculated using \(t := \tan(\alpha)\):

\(
\begin{align}
\sin(2 \alpha) &= \frac{2 t}{1 + t^2} \\
\\
\tan(2 \alpha) &= \frac{2 t}{1 - t^2} \\
\end{align}
\)

Solving the Quadratic Equation using \(\sin^{-1}\)

The first formula allows us to solve the following quadratic equation:

\(
t^2 - 2r \; t + 1 = 0
\)

We set \(r := \frac{1}{\sin(2 \alpha)}\) and rearrange the equation:

\(
\begin{align}
\frac{1}{r} = \frac{2 t}{1 + t^2} \\
\\
1 + t^2 = 2t \; r \\
\\
t^2 - 2r \; t + 1 = 0 \\
\end{align}
\)

So how can we use this to solve the quadratic equation: \(x^2 + px + q = 0\) ?
We assume \(q > 0\) so we can substitute: \(x = \sqrt{q} \; t\)
With this we get:

\(
\begin{align}
x^2 + px + q &= 0 \\
q t^2 + \sqrt{q} pt + q &= 0 \\
t^2 + \frac{p}{\sqrt{q}} t + 1 &= 0 \\
\end{align}
\)

We just have to set:

\(
\begin{align}
-2r &= \frac{p}{\sqrt{q}} \\
r &= - \frac{p}{2\sqrt{q}} \\
\end{align}
\)

But now we only have to stick everything together:

\(
\begin{align}
x
&= \sqrt{q} \; t \\
\\
&= \sqrt{q} \; \tan(\alpha) \\
\\
&= \sqrt{q} \; \tan\left(\frac{1}{2}\sin^{-1}\left(\sin(2 \alpha)\right)\right) \\
\\
&= \sqrt{q} \; \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{1}{r}\right)\right) \\
\\
&= \sqrt{q} \; \tan\left(\frac{1}{2}\sin^{-1}\left(- \frac{2\sqrt{q}}{p}\right)\right) \\
\end{align}
\)

Geometric Interpretation using the First Formula

There's a nice geometric interpretation of the first formula using the inscribed angle theorem:



I've already posted a similar picture in the previous post #36.

We were solving: \(x^2 - cx + h^2 = 0\)
Thus we'd end up with:

\(
\begin{align}
r
&= \frac{c}{2} \\
&= \frac{-p}{2} \\
\\
h^2 &= q \\
h &= \sqrt{q} \\
\\
\sin(2 \alpha)
&= \frac{h}{r} \\
&= -\frac{2 \sqrt{q}}{p} \\
\end{align}
\)

Solving the Quadratic Equation using \(\tan^{-1}\)

The second formula allows us to solve the following quadratic equation:

\(
t^2 + 2r \; t - 1 = 0
\)

We set \(r := \frac{1}{\tan(2 \alpha)}\) and rearrange the equation:

\(
\begin{align}
\frac{1}{r} = \frac{2 t}{1 - t^2} \\
\\
1 - t^2 = 2t \; r \\
\\
t^2 + 2r \; t - 1 = 0 \\
\end{align}
\)

So how can we use this to solve the quadratic equation: \(x^2 + px + q = 0\) ?
We assume \(q < 0\) so we can substitute: \(x = \sqrt{-q} \; t\)
With this we get:

\(
\begin{align}
x^2 + px + q &= 0 \\
-q t^2 + \sqrt{-q} pt + q &= 0 \\
t^2 + \frac{p}{\sqrt{-q}} t - 1 &= 0 \\
\end{align}
\)

We just have to set:

\(
\begin{align}
2r &= \frac{p}{\sqrt{-q}} \\
r &= \frac{p}{2\sqrt{-q}} \\
\end{align}
\)

But now we only have to stick everything together:

\(
\begin{align}
x
&= \sqrt{-q} \; t \\
\\
&= \sqrt{-q} \; \tan(\alpha) \\
\\
&= \sqrt{-q} \; \tan\left(\frac{1}{2}\tan^{-1}\left(\tan(2 \alpha)\right)\right) \\
\\
&= \sqrt{-q} \; \tan\left(\frac{1}{2}\tan^{-1}\left(\frac{1}{r}\right)\right) \\
\\
&= \sqrt{-q} \; \tan\left(\frac{1}{2}\tan^{-1}\left(\frac{2\sqrt{-q}}{p}\right)\right) \\
\end{align}
\)

Double Angle Formulae for Hyperbolic Functions

We get similar double angle formulae for the hyperbolic sine and hyperbolic tangent:

\(
\begin{align}
\sinh(2 \alpha) &= \frac{2 \tanh(\alpha)}{1 - \tanh(\alpha)^2} \\
\\
\tanh(2 \alpha) &= \frac{2 \tanh(\alpha)}{1 + \tanh(\alpha)^2} \\
\end{align}
\)

The astute reader may notice the change of the sign.
A similar calculation leads to the following formulae:

\(
\begin{align}
x &= \sqrt{-q} \; \tanh\left(\frac{1}{2}\sinh^{-1}\left(\frac{2\sqrt{-q}}{p}\right)\right) \\
\\
x &= \sqrt{q} \; \tanh\left(\frac{1}{2}\tanh^{-1}\left(- \frac{2\sqrt{q}}{p}\right)\right) \\
\end{align}
\)

Programs for the HP-42S

These programs are all similar. They calculate only one of the solutions.
If you want to use them with calculators that don't support complex numbers you may want to select the one that fits the sign of \(q\).

q > 0 using sin^-1:

Code:

00 { 20-Byte Prgm }
01▸LBL "Q"
02 SQRT
03 ÷
04 LASTX
05 X<>Y
06 -2
07 X<>Y
08 ÷
09 ASIN
10 2
11 ÷
12 TAN
13 ×
14 END

\(x^2 - 5x + 6 = 0\)

-5 ENTER 6 XEQ "Q"

2


q < 0 using tan^-1:

Code:

00 { 20-Byte Prgm }
01▸LBL "Q"
02 +/-
03 SQRT
04 ÷
05 LASTX
06 X<>Y
07 2
08 X<>Y
09 ÷
10 ATAN
11 2
12 ÷
13 TAN
14 ×
15 END

\(x^2 + x - 6 = 0\)

1 ENTER -6 XEQ "Q"

2


q > 0 using tanh^-1:

Code:

00 { 20-Byte Prgm }
01▸LBL "Q"
02 SQRT
03 ÷
04 LASTX
05 X<>Y
06 -2
07 X<>Y
08 ÷
09 ATANH
10 2
11 ÷
12 TANH
13 ×
14 END

\(x^2 - 5x + 6 = 0\)

-5 ENTER 6 XEQ "Q"

2


q < 0 using sinh^-1:

Code:

00 { 20-Byte Prgm }
01▸LBL "Q"
02 +/-
03 SQRT
04 ÷
05 LASTX
06 X<>Y
07 2
08 X<>Y
09 ÷
10 ASINH
11 2
12 ÷
13 TANH
14 ×
15 END

\(x^2 + x - 6 = 0\)

1 ENTER -6 XEQ "Q"

2


References

• Short Quadratic Solver (HP-42S): Message #32
  

[Albert Chan]

(05-12-2024 08:10 PM)Thomas Klemm Wrote:  \(
x = \sqrt{q} \; \tanh\left(\frac{1}{2}\tanh^{-1}\left(- \frac{2\sqrt{q}}{p}\right)\right)
\)

From any 1 formula, we can derived the other 3

tanh(2x) = 2*tanh(x) / (1 + tanh(x)^2)
sin(2x)   = 2*tan(x)   / (1 + tan(x)^2)

These have the same form, we get:

\(
x = \sqrt{q} \; \tan\left(\frac{1}{2}\sin^{-1}\left(- \frac{2\sqrt{q}}{p}\right)\right)
\)

Let z = GM/AM = √(q)/(-p/2), we "slide" ±i to the left, next to z

x = √(-q) * ±i * tan(asin(z)/2)        // √(q) = ±i * √(-q), sign to get LHS=RHS
   = √(-q) * tanh(±i * asin(z)/2)
   = √(-q) * tanh(asinh(±i*z)/2)      // ±i * √(q) = (±i)^2 * √(-q) = - √(-q)

\(
x = \sqrt{-q} \; \tanh\left(\frac{1}{2}\sinh^{-1}\left(\frac{2\sqrt{-q}}{p}\right)\right)
\)

Similarly, from quoted tanh/atanh formula, to tan/atan:

\(
x = \sqrt{-q} \; \tan\left(\frac{1}{2}\tan^{-1}\left(\frac{2\sqrt{-q}}{p}\right)\right)
\)

[7alken]

hi, if you want some mild puzzle, lets try to figure out what's happening here ... it took me several hours after trying it on canon f-915sg (just for fun, initially ...), when I absolutely didn't knew what I am doing nor the guy in the video ... but after that hours, I got more about the casio internals, about the quadratic solution, about the al-Khwarizmi, etc ... okay

But still don't know for sure if I will apologize to him or not ))

https://www.youtube.com/watch?v=Mi-49zfnfXs

it all was followed with gpt4o chat and then really several hours of investigation anywhere, what the hell is going on ...
https://chatgpt.com/share/0daa0ff3-8c20-...fae338c984

in school I hated math/history, now I quite love both, and always the programmable calculators, my first was Solatrex (Ireland) Calcul PSR-98e (citizen SR-59 like), when I could only dream about TI/HP from ham magazines of my dad ... now I have here also HP-12C Platinum and SwissMicro DM-15/16 and Prime and fx-CG50 (because they are top in hw), but fx-5800p is nice too; honestly, today, experimenting with repurposing some calculators hw (keyb/case) for something different inside and outside (ips or amoled display)...

(my honest oppinion is that scientific calc and his thick manual is best way to (re)learn math basics; memorizing of any solution is of course nonsene... these days; understanding is crucial)

cheers
7alken

BTW: already really apologized to him, as on casio fx-82ms (2nd) it REALLY works )) ... on canon f-715sg no ... so, its still a puzzle; no more time to investigate more ...

[Albert Chan]

Hi, 7alken

Welcome to the forum!

The trick to force quadratic regression line == quadratic equation to solve. (*)
Let N = number of data points (N=0 when calculator memory is cleared)

{N,f(N)}, entered 3 times, we have points {0,f(0)}, {1,f(1)}, {2,f(2)}

If f is quadratic, quadratic regression line = f --> regression line roots = f roots.

For casio fx-82ms (2nd), variable C=N, thus {C,F(C)} work.
For other models, you may need to figure out what variable stored N

(*) your video link example, f(x) = 112*x^2 + 100*x - 525
If you checked regression lines, you'll see C, B, A = 112, 100, -525
Points are used to indirectly enter these coefficients.