Solve my determinant equal to zero with roots

Hi!

I need help with solving for the roots of my polynomial im getting from my determinant of my matrices in a way where i dont have to collect the values manually in front of the variabel P from the determinant. Thanks in advance!

%%
clear all
clc
syms P

% Matrix definitions
KE_red =[13.0000         0   -8.4853;
    0   13.0000         0;
    -8.4853         0    8.0000];

KG_red =[ -(3*2^(1/2)*P)/65,   0,    P/130;
    0, -(3*2^(1/2)*P)/65, 0
    P/130,  0, -(2*2^(1/2)*P)/195];

% Determinant       
D=det(KE_red + KG_red)

% Solve for the roots of the determinant = 0
p = [-9*2^(1/2)/219700, 167/3380, -1328*2^(1/2)/195, 416];
r = roots(p)

[SL: formatted code as code. In the future please use the first button in the Code section of the toolstrip in the MATLAB Answers editor to create a section that formats code as code.]

Answer 1

[Steven Lord]

You can use the vpasolve function to solve the polynomial, use the sym2poly function to automate extracting the vector of coefficients to a double precision vector, or use the coeffs function to automate extracting the vector of coefficients to a symbolic vector.

%%
syms P
sympref('FloatingPointOutput', false);

% Matrix definitions
KE_red =[13.0000         0   -8.4853;
    0   13.0000         0;
    -8.4853         0    8.0000];

KG_red =[ -(3*2^(1/2)*P)/65,   0,    P/130;
    0, -(3*2^(1/2)*P)/65, 0
    P/130,  0, -(2*2^(1/2)*P)/195];

% Determinant       
D=det(KE_red + KG_red)

D = 

% Solve for the roots of the determinant = 0
p = [-9*2^(1/2)/219700, 167/3380, -1328*2^(1/2)/195, 416];
r = roots(p)

r = 3x1

592.8693 199.1684 60.8116

s = vpasolve(D)

s = 

p2 = sym2poly(D)

p2 = 1x4

-0.0001 0.0494 -9.6311 415.9959

r2 = roots(p2)

r2 = 3x1

592.8696 199.1684 60.8110

p3 = coeffs(D, 'all') % Handle the case where one of the powers is not present

p3 = 

r3 = roots(p3)

r3 = 

To check, let's sort the various vectors. I'll use vpa to approximate the symbolic answers.

vpa([sort(r), sort(s), sort(r2), sort(r3)], 8)

ans = 

Those look to be in pretty close agreement.

[Daniel]

Thank you @Steven Lord