new way to make quadratic equations easy
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07-30-2021, 01:44 AM
Post: #1
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new way to make quadratic equations easy
Thought this might be of interest here..
https://getpocket.com/explore/item/mathe...-equations https://www.technologyreview.com/2019/12...ions-easy/ https://www.poshenloh.com/quadratic/ |
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07-30-2021, 11:42 AM
Post: #2
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RE: new way to make quadratic equations easy
Hello!
Simpler? I watched the video. I am not convinced. And what if the coefficients are not integers? Regards Max |
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07-30-2021, 01:46 PM
Post: #3
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RE: new way to make quadratic equations easy
(07-30-2021 11:42 AM)Maximilian Hohmann Wrote: Hello! If they are not integers, I give up! (I was/am not good at quadratics, it is sort of the reason I hang around this forum like a rock star groupie, I am so awed by people able to find i or some other negative number as the root!) 10B, 10BII, 12C, 14B, 15C, 16C, 17B, 18C, 19BII, 20b, 22, 29C, 35, 38G, 39G, 41CV, 48G, 97 |
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07-31-2021, 04:22 AM
(This post was last modified: 07-31-2021 04:22 AM by Namir.)
Post: #4
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RE: new way to make quadratic equations easy
Seems to work with non-integers. I wrote an HP-41CX program that takes the roots, forms the coefficients B and C of the quadratic equation, and then solves it. Input and output values match!
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07-31-2021, 07:03 AM
Post: #5
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RE: new way to make quadratic equations easy
Here is an HP41/42 Implementation using only the stack. The stack starts with C in register Y and B in register X: The two roots will be in registers X and Y.
Code:
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07-31-2021, 10:19 AM
(This post was last modified: 07-31-2021 10:47 AM by C.Ret.)
Post: #6
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RE: new way to make quadratic equations easy
(07-31-2021 04:22 AM)Namir Wrote: Seems to work with non-integers. It also works for complex roots or complex coefficients, here is an illustration for a complex capable HP-15C: Code: IN : Y: c X: b from reduced quadratic equation x²+b.x+c = 0 (EQ.1) Usage : Enter real or complex reduced coefficients c and b in stack at respectively level Y: and X: and run the code. HP-15C 's complex mode automatically engage at complex value entries or when any of the two roots is complex. This code, despite using a different method is really close to the one I post yesterday using old-school method. Usage Example : \( (3-i).z^2+(-49+33i).z+120-90i = 0 \) Enter c coefficient : [ 1 ][ 2 ][ 0 ][ENTER] [ 9 ][ 0 ][CHS] f[ I ] (note that the 'c' annunciators lit on) Reduce c by a : [ 3 ][ENTER] [ 1 ][CHS] f[ I ] [ ÷ ] display reduced c : " 45.00000 c " Press f[(i)] to see imaginary part "-15.00000 c" Enter b: [ 4 ][ 9 ][CHS][ENTER] [ 3 ][ 3 ] f[ I ] Reduce b by a : g[LSTx] [ [ ÷ ] display reduced b : "-18.00000 c " Press f[(i)] to see imaginary part " 5.0000" The reduced quadratic equation is now : \( z^2+(-18+5i).z+(45-15i)=0 \) Press [R/S] to run code. Real part of first root is " 15.00000 " Press and hold f[(i)] key to see imaginary part of first root: "-5.00000 " Press [X↔Y] to display the second root real part : " 3.00000 " Press and hold f[(i)] to display the second root’s imaginary part: " 0.00000 " The quadratic equation \( (3-i).z^2+(-49+33i).z+120-90i = 0 \) have one complex root \(z_1=15-5i\) and one real root \(x_2=3 \) |
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08-01-2021, 07:56 AM
Post: #7
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RE: new way to make quadratic equations easy
Updated version (using stack only) to handle real and complex solutions for real values of B and C.
Code: LBL A If flag 00 remains clear at the end of the calculations you have two real roots in the X and Y registers. If flag 00 is set at the end of the calculations thenn you have the solutions: x1 = regX + i regY x2 = regX - i regY Namir |
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08-01-2021, 03:08 PM
(This post was last modified: 08-01-2021 07:55 PM by Albert Chan.)
Post: #8
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RE: new way to make quadratic equations easy
(07-31-2021 10:19 AM)C.Ret Wrote: ... The reduced quadratic equation is now : \( z^2+(-18+5i).z+(45-15i)=0 \) We can also solve quadratics with half-angle formula, see (HP-67) Barkers's Equation Let c = cot(θ/2) → cot(θ) = (c²-1) / (2c) → c² - 2*cot(θ)*c - 1 = 0 x² - 2*cot(θ)*x - 1 = (x - c)*(x + 1/c) Let x = z/n, to scale constant term to -n², instead of -1. In other words, solve for z² - 2*m*z - n² = 0 XCas> m,n := (-18.+5i)/-2, sqrt(-(45.-15i)) XCas> proot([1, -2m, -n*n]) → [3.0, 15.0-5.0*i] XCas> -n*tan(atan(n/m)/2), n/tan(atan(n/m)/2) → [3.0, 15.0-5.0*i] Or, with "built-in" quadratic solver, asinh: sinh(x) = (e^x - e^-x)/2 → (e^x)² - 2*sinh(x)*e^x - 1 = 0 XCas> -n/exp(asinh(m/n)), n*exp(asinh(m/n)) → [3.0, 15.0-5.0*i] Since asinh(x) is odd function, z = ±n*exp(asinh(m/±n)) --- We can also solve for z² - 2*m*z + n² = 0 XCas> m,n := (-18+5i)/-2., sqrt(45.-15i) XCas> proot([1, -2m, n*n]) → [3.0, 15.0-5.0*i] XCas> n*tan(asin(n/m)/2), n/tan(asin(n/m)/2) → [3.0, 15.0-5.0*i] XCas> n/exp(acosh(m/n)), n*exp(acosh(m/n)) → [3.0, 15.0-5.0*i] Again, combine both roots, z = n*exp(±acosh(m/n)) |
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08-04-2021, 04:57 AM
Post: #9
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RE: new way to make quadratic equations easy
I was surprised to learn that roots for quadratic equations could be solved using a slide rule, described in the manual for the Post Versalog:
Quote:If any quadratic equation is transformed into the form x^2 + Ax + B = 0, the roots or values of the unknown x may be determined by a simple method, using the slide rule scales. We let the correct roots be -x_1 and -x_2. By factoring, (x + x_1)(x + x_2) = 0. The terms -x_1 and -x_2 will be the correct values of x providing the sum x_1 + x_2 = A and the product of x_1 * x_2 = B. An index of the CI scale may be set opposite the number B on the D scale. With the slide in this position, no matter where the hairline is set, the product of simultaneous CI and D scale readings or of simultaneous CIF and DF scale readings is equal to B. Therefore it is only necessary to move the hairline to a position such that the sum of the simultaneous CI and D scale readings, or the sum of the simultaneous CIF and DF scale readings, is equal to the number A. |
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08-06-2021, 12:40 AM
(This post was last modified: 08-06-2021 03:36 AM by Namir.)
Post: #10
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RE: new way to make quadratic equations easy
Interseting article. I do have a "small" collection of slide rules. I used them in high school and during the first part of my engineering education until I got a Sharp scientific calculator and soon after than an HP-55 (in 1975).
My favorite slide rule is a big Aristo (with a two-piece support stand) that my father bought me from Lausanne, Switzerland, in 1971. I only brought it to final exams. I never brought the HP-55 to college. I think slide rules are amazing in their own right. But, the HP-35 ushered their demise. Namir |
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08-06-2021, 02:22 PM
Post: #11
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RE: new way to make quadratic equations easy
(08-06-2021 12:40 AM)Namir Wrote: Interseting article. I do have a "small" collection of slide rules. [...] I think my "collection" is <30, it includes a concrete calculator given to customers of a Ready-Mix company, and an Addiator, oh, and a paper flight slide rule (Jeppeson?) I did buy a non-functioning slide rule tie tack on TAS a while back, so I don't think that counts. 10B, 10BII, 12C, 14B, 15C, 16C, 17B, 18C, 19BII, 20b, 22, 29C, 35, 38G, 39G, 41CV, 48G, 97 |
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08-06-2021, 05:32 PM
(This post was last modified: 08-07-2021 09:49 AM by C.Ret.)
Post: #12
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RE: new way to make quadratic equations easy
(08-04-2021 04:57 AM)Benjer Wrote: I was surprised to learn that roots for quadratic equations could be solved using a slide rule, described in the manual for the Post Versalog: I was also really surprise, so I grasp my father's Graphoplex REITZ n°620 and it's very short manual where no mention about how to solve a quadratic equation is present. So I reinserted this manual in the red cardboard box and try the two examples. Here is a capture for the resolution of the first example : \( x^2+10x+15=0 \) . The trick is to mentally add value form the CI and the D scale. I get a hard time to found at which position I have to start seeking. To help me, I move the hairline (the middle hairline in red) to look for a sum as close as 10 as possible. Finally, the position of the middle hairline indicate the two roots on the CI and D scale respectively. [attachment=9708] (Note: the HP-15C armed with quadratic solving code is only showing one root at a time). For the second example of equation \( x^2–12.2x-17.2=0 \), I check with an HP Prime to have the two roots on the display. I only realize after a few attempts, that since there is no CIF and DF scale on this slide rule I need to use the right index of CI set on 17.2 on the D scale. Then, the procedure is usingthe D and CI scales but the substraction have to match 12.2 as close as possible. [attachment=9709] P.S.: Please note that on this slide rule, scale CI and D have respectively black B and red a labels (original Graphoplex: French Règle à Calculs). EDIT: Is that a new way to make the quadratic equation solving esay ? - new ? With a slide rule from 1958 ? - easy ? scanning for adding or substrating values on D and CI scale ? |
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