**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

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