How can I make a loop for solving a quadratic equation with 200 different data points and only include the positive values of the quadratic eqaution?

p1 = RelaxKineticsData.pH0_2

p1e = 6.7240 p1f = p1 - p1e c1H = 10.^(-p1f) t

1 = RelaxKineticsData.Time0_2

Ka = 0.00000074 Kd = 50

Cr1 = 0.086

syms x eqn = (Ka*x)/c1H + x + Kd*x^2 - 0.086 == 0 double(solve(eqn,x)) So far this is what I have.

I also tried p1 = RelaxKineticsData.pH0_2;

p1e = 6.7240;

p1f = p1 - p1e;

c1H_values = 10.^(-p1f);

t1_values = RelaxKineticsData.Time0_2;

Ka = 0.00000074;

Kd = 50;

Cr1 = 0.086;

x_values = zeros(size(c1H_values));

for i = 1:length(c1H_values)

c1H = c1H_values(i);

eqn = (Ka*x)/c1H + x + Kd*x^2 - Cr1 == 0;

x_values(i) = double(solve(eqn, x));

end

and got this error : Unable to perform assignment because the left and right sides have a different number of elements. Error in untitled2 (line 20) x_values(i) = double(solve(eqn, x));

This seems like a rather simple task, just calculating positive values of a quadratic equation for a set of data, but It's been giving me alot of trouble.

Answer 1

[Dyuman Joshi]

"eqn" is a quadratic equation in x and thus using solve() will give you 2 outputs. But you are trying to assign 2 outputs to one numeric place holder, which gives you the error.

n = numel(c1H_values);
%Initiate x_values as a two column array to
%store both roots of the equation
x_values = zeros(n,2);
for i = 1:n
    c1H = c1H_values(i);
    eqn = (Ka*x)/c1H + x + Kd*x^2 - Cr1 == 0;
    x_values(i,:) = double(solve(eqn, x));
end

You can set non-positive roots of the equations to 0 or NaN.

In case you just want to obtain the positive roots, define x_values as a cell array and proceed further.

[Ethan]

How can I define x_values as a cell array to only obtain the positive roots?

Thanks so much for your help!

[Dyuman Joshi]

n = numel(c1H_values);
%Initiate x_values as a cell array
%store positive roots of the equation
x_values = cell(n,1);
for k = 1:n
    c1H = c1H_values(k);
    eqn = (Ka*x)/c1H + x + Kd*x^2 - Cr1 == 0;
    temp = double(solve(eqn, x));
    %In case both the roots are imaginary or have a negative real component
    %the corresponding cell element will be empty
    x_values{k} = temp(real(temp)>0 & abs(imag(temp))<1e-10);
    %using tolerance instead of equality as we are dealing in floating
    %point number, change tolerance as per need
end

[Ethan]

When I tried to run the code, I got an error fairly quickly.

Unrecognized function or variable 'c1H_values'.

Error in untitled2 (line 11)

n = numel(c1H_values)

How can I define c1H_values?

[Dyuman Joshi]

Like you did earlier -

p1 = RelaxKineticsData.pH0_2;
p1e = 6.7240;
p1f = p1 - p1e;
c1H_values = 10.^(-p1f);

I only showed a part of your code.