# How to know results

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Ani Asoyan on 2 Jun 2020
Commented: Ameer Hamza on 3 Jun 2020
I have a polynomial and parameters in it,
a=2, c=1, d=2 , g=0.5, k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]
p=[ -c.*k+ (f.^2)*(e.^2)/a+2*e.*g, d.*e-(f*b.*e)/a]
roots(p)
I calculate roots of polynomial given parameter values.
But e an b parameters have vector values.
I want to know if it obtains roots for every combination of the e and b values or
it simply gets roots when e=2, b=1 , then b=8,e=2,... ?
##### 1 CommentShowHide None
Ani Asoyan on 2 Jun 2020
I want to know if it takes into account cross-combinations of e and b ?

Ameer Hamza on 2 Jun 2020
Edited: Ameer Hamza on 2 Jun 2020
Your current code does not solve the problem, as you described. You need to use for-loop
a=2, c=1, d=2 , g=0.5, k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]
r = zeros(numel(e), 1) % polynomial is 1st order so there will be only one root
for i=1:numel(e)
p = [ -c.*k+(f.^2)*(e(i).^2)/a+2*e(i).*g, d.*e(i)-(f*b(i).*e(i))/a]
r(i) = roots(p)
end
Ameer Hamza on 3 Jun 2020
Following code use cross-combination
a=2, c=1, d=2 , g=0.5, k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]
[E, B] = ndgrid(e, b);
e = E(:);
b = B(:);
r = zeros(numel(e), 1) % polynomial is 1st order so there will be only one root
for i=1:numel(e)
p = [ -c.*k+(f.^2)*(e(i).^2)/a+2*e(i).*g, d.*e(i)-(f*b(i).*e(i))/a]
r(i) = roots(p)
end
sol = [e, b, r]
1st column of 'sol' is 'e' value, 2nd is 'b' values, and 3rd is the corresponding root.

### More Answers (1)

Steven Lord on 2 Jun 2020
As written your code solves for the roots of the 7th order polynomial whose coefficients are:
p =
8 134 53 34 2 0 -5 -8
This is:
8*x^7 + 134*x^6 + 53*x^5 + 34*x^4 + 2*x^3 - 5*x - 8
If instead you want to find the roots of the four first order polynomials created by specifying each element of e and b in the expression for p, use a for loop instead.
>> for q = 1:numel(e)
p=[ -c.*k+ (f.^2)*(e(q).^2)/a+2*e(q).*g, d.*e(q)-(f*b(q).*e(q))/a]
end
p =
8 2
p =
134 0
p =
53 -5
p =
34 -8
If you want to find the roots of the sixteen first order polynomials with elements taken from e and from b (and not necessarily the same element of each vector) the easiest way is to use a double for loop.
##### 2 CommentsShowHide 1 older comment
Steven Lord on 2 Jun 2020
It is not first order as you've written it. Use one or two for loops to construct and solve the four or sixteen first order polynomials or use the explicit formula for the solution of a linear equation.
% if q*x = w then x = w./q
a=2, c=1, d=2 , g=0.5, k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]
q = -c.*k+ (f.^2)*(e.^2)/a+2*e.*g;
w = d.*e-(f*b.*e)/a;
x = w./q % Using ./ instead of / because w and q are vectors
check = q.*x - w % Should be all 0's or close to it