new way to make quadratic equations easy

07302021, 01:44 AM
Post: #1




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...ionseasy/ https://www.poshenloh.com/quadratic/ 

07302021, 11:42 AM
Post: #2




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 

07302021, 01:46 PM
Post: #3




RE: new way to make quadratic equations easy
(07302021 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, 41CV, 48G, 97 

07312021, 04:22 AM
(This post was last modified: 07312021 04:22 AM by Namir.)
Post: #4




RE: new way to make quadratic equations easy
Seems to work with nonintegers. I wrote an HP41CX program that takes the roots, forms the coefficients B and C of the quadratic equation, and then solves it. Input and output values match!


07312021, 07:03 AM
Post: #5




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:


07312021, 10:19 AM
(This post was last modified: 07312021 10:47 AM by C.Ret.)
Post: #6




RE: new way to make quadratic equations easy
(07312021 04:22 AM)Namir Wrote: Seems to work with nonintegers. It also works for complex roots or complex coefficients, here is an illustration for a complex capable HP15C: 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. HP15C '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 oldschool method. Usage Example : \( (3i).z^2+(49+33i).z+12090i = 0 \) 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 : \( z^2+(18+5i).z+(4515i)=0 \) 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 \( (3i).z^2+(49+33i).z+12090i = 0 \) have one complex root \(z_1=155i\) and one real root \(x_2=3 \) 

08012021, 07:56 AM
Post: #7




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 

08012021, 03:08 PM
(This post was last modified: 08012021 07:55 PM by Albert Chan.)
Post: #8




RE: new way to make quadratic equations easy
(07312021 10:19 AM)C.Ret Wrote: ... The reduced quadratic equation is now : \( z^2+(18+5i).z+(4515i)=0 \) We can also solve quadratics with halfangle formula, see (HP67) Barkers's Equation Let c = cot(θ/2) → cot(θ) = (c²1) / (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, 15.05.0*i] XCas> n*tan(atan(n/m)/2), n/tan(atan(n/m)/2) → [3.0, 15.05.0*i] Or, with "builtin" 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, 15.05.0*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, 15.05.0*i] XCas> n*tan(asin(n/m)/2), n/tan(asin(n/m)/2) → [3.0, 15.05.0*i] XCas> n/exp(acosh(m/n)), n*exp(acosh(m/n)) → [3.0, 15.05.0*i] Again, combine both roots, z = n*exp(±acosh(m/n)) 

08042021, 04:57 AM
Post: #9




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. 

08062021, 12:40 AM
(This post was last modified: 08062021 03:36 AM by Namir.)
Post: #10




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 HP55 (in 1975).
My favorite slide rule is a big Aristo (with a twopiece support stand) that my father bought me from Lausanne, Switzerland, in 1971. I only brought it to final exams. I never brought the HP55 to college. I think slide rules are amazing in their own right. But, the HP35 ushered their demise. Namir 

08062021, 02:22 PM
Post: #11




RE: new way to make quadratic equations easy
(08062021 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 ReadyMix company, and an Addiator, oh, and a paper flight slide rule (Jeppeson?) I did buy a nonfunctioning 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, 41CV, 48G, 97 

08062021, 05:32 PM
(This post was last modified: 08072021 09:49 AM by C.Ret.)
Post: #12




RE: new way to make quadratic equations easy
(08042021 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. (Note: the HP15C armed with quadratic solving code is only showing one root at a time). For the second example of equation \( x^2–12.2x17.2=0 \), 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. 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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