Solve a quadratic equation

4 ビュー (過去 30 日間)
Mepe
Mepe 2020 年 2 月 20 日
コメント済み: Mepe 2020 年 2 月 20 日
So far I have solved the equation below with fsolve (with the help of this forum).
tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
eq = @(s,f4,f8) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
for f = 1:1:length (f4)
F1 (f,:) = fsolve (@(s)eq(s,f4(f),f8(f)), 0);
end
Unfortunately, only a solution of the quadratic equation is given here. I didn't get along with the command roots () because my "formulas" were not accepted here. Does anyone have an idea here how elegantly all solutions can be found?

採用された回答

Alex Mcaulley
Alex Mcaulley 2020 年 2 月 20 日
編集済み: Alex Mcaulley 2020 年 2 月 20 日
To use the function roots you need to reformulate your equation:
tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
eq = @(f4,f8) [-0.1*f4^2, -3.54*f4^2 + tau,9.53*f4^2-f8];
sol = zeros(numel(f4),2);
for f = 1:1:length(f4)
sol(f,:) = roots(eq(f4(f),f8(f)));
end
>> sol
sol =
-37.7542 2.4654
-37.3027 2.1527
-37.8394 2.4672
-37.8874 2.5030
  1 件のコメント
Mepe
Mepe 2020 年 2 月 20 日
Works perfect. Thanks!

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangePartial Differential Equation Toolbox についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by