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+--- Thread: (17BII and 27S) Quadratic Equation (/thread-12686.html)
(17BII and 27S) Quadratic Equation - Eddie W. Shore - 03-26-2019 04:05 PM
The following solver equations solve the quadratic equation
A*x^2 + B*x + C = 0
by the famous Quadratic Formula
x = (-B ± √(B^2 - 4*A*C) ) / (2*A)
Define D as the discriminant: D = B^2 - 4*A*C
If A, B, and C are real numbers and:
D<0, the roots are complex conjugates
D≥0, the roots are real roots
Quadratic Equation: Real Roots Only
Code:
QUAD:X=INV(2*A)*(-B+SQRT(B^2-4*A*C)*SGN(R#))Input Variables:
A: coefficient of X^2
B: coefficient of X
C: constant
R#: -1 or 1
Output Varibles:
X: root
Example: 2X^2 + 3X - 5 = 0
Input:
A: 2
B: 3
C: -5
R#: 1 (or any positive number)
Output:
X = 1
Input:
R#: -1
Output:
X = -2.5
Quadratic Equation: Real or Complex Roots
(Let (L) and Get (G) functions required)
Code:
QUAD:0*(A+B+C+L(D:B^2-4*A*C)+L(E:2*A))
+IF(S(X1):IF(D<0:-B÷G(E):(-B+SQRT(D))÷G(E))-X1:0)
+IF(S(X2):IF(D<0:SQRT(ABS(D))÷G(E):(-B-SQRT(D))÷G(E))-X2:0)Input Variables:
A: coefficient of X^2
B: coefficient of X
C: constant
Output Variables:
D: Discriminant
If D<0: X1: real part, X2: imaginary part
If D≥0: X1: real root 1, X2: real root 2
Example 1: -3*X^2 + 8*X - 1= 0
Input:
A: -3
B: 8
C: -1
Output:
D = 52
X1 = 0.1315
X2 = 2.5352
Roots: x = 0.1315, x = 2.5352
Example 2: 3*X^2 + 5*X + 3 = 0
Input:
A: 3
B: 5
C: 3
Output:
D = -11
X1 = -0.8333
X2 = 0.5528
Roots: x = -0.8333 ± 0.5528i
Link: https://edspi31415.blogspot.com/2019/03/hp-17bii-and-hp-27s-quadratic-formula.html