TI 30Xa Algorithms: Solving Monic Quadratic Polynomials Quickly
We all know about the tried and true Quadratic Formula to solve quadratic equations. However, it is not the only way to tackle such problems.
Today’s blog is covers a way to quickly get the roots of the quadratic equation:
x^2 + p * x + q = 0
I am focusing on monic quadratic polynomials today. A polynomial is a monic polynomial if the leading coefficient is 1. In this instance, the coefficient of the x^2 term is 1. I’m also going to work with quadratic equations that have real roots.
Deviation
Please see the source article as the derivation and method is explained by the author Po-Shen Loh (https://www.poshenloh.com/quadraticdetail/). This method has also been developed by Viete and other classic mathematicians. Here I attempt to explain a derivation of this method.
Let r1 and r2 be the two roots of the polynomial and x^2 + p * x + q factors to:
x^2 + p * x + q = (x – r1) * (x – r2)
Expanding (x – r1) * (x – r2):
(x – r1) * (x – r2) = x^2 + (-r1 -r2) * x + r1 * r2
Then:
p = - r1 – r2 ⇒ r1 + r2 = -p
q = r1 * r2
Let one of the roots be defined as: [see Source]
r1 = -p/2 + u
Then:
r1 + r2 = -p
-p/2 + u + r2 = -p
r2 = -p/2 + u
And:
r1 * r2 = q
(-p/2 + u) * (-p/2 - u) = q
p^2/4 – u^2 = q
- u^2 = q – p^2/4
u^2 = p^2/4 – q
u = ±√(p^2/4 – q)
And the roots are:
r1 = -p/2 + √(p^2/4 – q)
r2 = -p/2 - √(p^2/4 – q)
x = -p/2 ± u
Note:
r1 = -p/2 + u
r1 – 2 * u = -p/2 + u – 2 * u
r1 – 2 * u = -p/2 – u
r1 – 2* u = r2
We can verify the above result with the quadratic equation:
x = (-p ± √( p^2 – 4 * q)) / 2
x = -p/2 ± √( p^2 – 4 * q) / 2
x = -p/2 ± √(( p^2 – 4 * q) / 4)
x = -p/2 ± √(( p^2/4 – q)
x = -p/2 ± u
TI-30Xa Algorithm: Quadratic Equation
Assumption: The roots are real (not complex).
Keystrokes:
p [ STO ] 1
q [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (r1, root 1)
[ ± ] [ - ] [ RCL ] 1 [ = ] (r2, root 2)
Memory registers used: M1 = p, M2 = q
Examples
Example 1: x^2 – 3*x – 4 = 0
p = -3, r = -4
3 [ ± ] [ STO ] 1
4 [ ± ] [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 2.5)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = 4)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = -1)
x = 4, -1
Example 2: x^2 – 24*x + 135 = 0
p = -24, r = 135
24 [ ± ] [ STO ] 1
135 [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 3)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = 15)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = 9)
x = 15, 9
Example 3: x^2 + 10*x + 24 = 0
p = 10, r = 24
10 [ STO ] 1
24 [ STO ] 2
[ ( ] [ RCL ] 1 [ x^2 ] [ ÷ ] 4 [ - ] [ RCL ] 2 [ ) ] [ √ ] (u = 1)
[ - ] [ RCL ] 1 [ ÷ ] 2 [ = ] (root 1, x = -4)
[ ± ] [ - ] [ RCL ] 1 [ = ] (root 2, x = -6)
x = -4, -6
Source
Loh, Po-Shen. “Quadratic Method: Detailed Explanation” Updated August 6, 2021. Retrieved May 19, 2024. https://www.poshenloh.com/quadraticdetail/
Until next time, have a great day. For the Americans, have a safe and sane Fourth of July. See you July 6!
Eddie
All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
