# I need to solve an equation involving squares and cubes.

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Thomas McCormack on 12 Nov 2019
I need to sole this equation for L, and I have a set of values for d and p.
p = -2(d/L)^3 + 3(d/L)^2
Im wondering how to put this into Matlab. Any help would be apprecitated.

#### 1 Comment

James Tursa on 12 Nov 2019
Multiply both sides by L^3 and then you have a polynomial in L to work with.

Dimitris Kalogiros on 12 Nov 2019
Edited: Dimitris Kalogiros on 13 Nov 2019
It is a 3rd order equation, it has only an analytical solution.
You can use the following code:
clc; clearvars;
syms p d L
p=1;
d=3;
eq= p == -2*(d/L)^3 + 3*(d/L)^2
% analytical solution
solve(eq,L)
%numerical solution
vpasolve(eq)
And in case you want to run this for a set of values , you can do the following:
clc; clearvars;
syms p d L
% values for p, d
pValues=50:10:90;
dValues=300:50:600;
% an empty buffer for the solutions, combined with p,d
SolBuffer=cell(length(pValues)*length(dValues),3);
% solution counter
k=0;
for n=1:length(pValues)
for m=1:length(dValues)
k=k+1;
p=pValues(n);
SolBuffer{k,1}=p;
d=dValues(m);
SolBuffer{k,2}=d;
eq= p == -2*(d/L)^3 + 3*(d/L)^2;
% analytical solution
solve(eq, L);
%numerical solution
SolBuffer{k,3}=vpasolve(eq).';
end
end
% An example: the third solution
disp('data p and d :')
disp(SolBuffer{3,1})
disp(SolBuffer{3,2})
disp('solutions:')
disp(SolBuffer{3,3})

Dimitris Kalogiros on 13 Nov 2019
Thomas McCormack on 13 Nov 2019
Thats very kind of you, thanks a lot!! One final thing, is there a way to print all the answers in a matrix?
Dimitris Kalogiros on 13 Nov 2019
clc; clearvars;
syms p d L
% values for p, d
pValues=50:10:90;
dValues=300:50:600;
% an empty buffer for the solutions, combined with p,d
SolBuffer=nan(length(pValues)*length(dValues),3);
% solution counter
k=0;
for n=1:length(pValues)
for m=1:length(dValues)
k=k+1;
p=pValues(n);
d=dValues(m);
eq= p == -2*(d/L)^3 + 3*(d/L)^2;
% analytical solution
solve(eq, L);
%numerical solution
SolBuffer(k,1:3)=vpasolve(eq).';
end
end
% display solutions
SolBuffer

John D'Errico on 12 Nov 2019
You have many values for both d and p? Do you want to solve it for all combinations of d and p?
Anyway, first, I would recognize that d and L are alwways combined in the same form, thus, we have d/L always. So first, substitute
X = d/L
Of course, once we know the value of X, we can always recover L, as:
L = d/X
p = -2*X^3 + 3*X^2
You could use a loop over all values of p, getting three roots for each value of p. Before we do even that however, do a plot.
ALWAYS PLOT EVERYTHING.
fun = @(X) -2*X.^3 + 3*X.^2;
fplot(fun,[-1,2])
yline(0);
yline(1);
See that I used the .^ operator there to avoid problems. Because this is a cubic polynomial in X, you can think of your problem as having 3 real roots for X, whenever p is betweeen 0 and 1. For p larger than 1 or smaller than 0, the problem will have one real root and two complex roots.
For example,
polycoef = @(p) [-2 3 0 -p];
roots(polycoef(0))
ans =
0
0
1.5
roots(polycoef(-.00001))
ans =
1.5 + 0i
-1.1111e-06 + 0.0018257i
-1.1111e-06 - 0.0018257i
So if p is just slightly less than zero as I predicted, we see a real root, but then always two complex conjugate complex roots. The same thing happens at p==1, where 3 real roots turn into a real and two complex roots as we go above p=1.
For p between 0 and 1 however, three real roots for X.
roots(polycoef(0.5))
ans =
1.366
0.5
-0.36603
Again, if you then know the value of d, we can easily recover the value of L, as
L = d./X