From symbolic to numerical results for quadratic equation
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
laura bagnale
am 17 Jun. 2021
Bearbeitet: Walter Roberson
am 19 Jun. 2021
Hello everyone,
I hope someone could help me, I'm starting using the symbolic toolbox but I have some difficulties.
I wrote this quadratic equation in symbolic form:
syms x y a1 b1 c1 d1 e1 f1
a = d1*y^2 + (e1 + b1*x)*y + a1*x^2 + c1*x + f1
then I tried to solve it:
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y)
I got what I expected from the formula
[-(e1 + b1*x - sqrt(b1^2*x^2 + 2*b1*e1*x + e1^2 - 4*a1*d1*x^2 - 4*c1*d1*x - 4*d1*f1))/(2*d1); -(e1 + b1*x + sqrt(b1^2*x^2 + 2*b1*e1*x + e1^2 - 4*a1*d1*x^2 - 4*c1*d1*x - 4*d1*f1))/(2*d1)]
However I don't know how to get the numerical real solutions.
I know the numerical values for the coeffcients a1, b1, c1, d1, e1, f1.
I tried to write the equation in this way:
syms x y
a = d1*y^2 + (e1 + b1*x)*y + a1*x^2 + c1*x + f1
ht=matlabFunction(a)
eqn = d1.*y.^2+(e1+b1.*x).*y+(a1.*x.^2+c1.*x+f1) == 0
S = solve(eqn, y, 'Real', true)
but the formula appears in the same form, with numbers in the places of the coefficients but with no solutions.
Can someone help me please.
Thank you very much in advance.
Laura
2 Kommentare
KSSV
am 17 Jun. 2021
Read about subs. Did you define the values of those coefficients? Show us the full code which you tried.
Akzeptierte Antwort
Stephan
am 17 Jun. 2021
Bearbeitet: Stephan
am 17 Jun. 2021
You could use symbolic functions:
syms y(a,b,x)
y(a,b,x) = a*x^2 - b*x
% Calculate values for a=1, b=2 and x in a range from 1:4 with stepwide 0.5
result_symbolic = y(1,2,1:0.5:4)
if you want that numbers (still symbolic) use double
result_numeric = double(result_symbolic)
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Special Values finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!