How can i solve this equation with matlab ?

x=-(2*10^4)*k
y=(2*10^4)*k
z=-3+1*(sqrt(y+0.6)-sqrt(0.6))
k=(10^(-4))*(x-z)^2

How can i find, x,y,z and k with matlab?

Answer 1

[Star Strider]

If you have the Symbolic Math Toolbox (with a slightly revised version of your equations) this works:

syms x y z k 
Eq1 = x == -2E4*k;
Eq2 = y == 2E4*k;
Eq3 = z == -3+(sqrt(y+0.6)-sqrt(0.6));
Eq4 = k == 1E-4*(x-z)^2;
[x,y,z,k] = solve(Eq1, Eq2, Eq3, Eq4, [x y z k])

To get one set of numerical solutions, use vpasolve, and explore the solution space.

See the documentation on solve and vpasolve for details.

[Duygu Demirpençe]

I'm new matlab, i have this equation for your codes what does it means?

x =

 3/5 - root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 1)^2
 3/5 - root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 2)^2
 3/5 - root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 3)^2
 3/5 - root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 4)^2

y =

 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 1)^2 - 3/5
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 2)^2 - 3/5
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 3)^2 - 3/5
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 4)^2 - 3/5

z =

 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 1) - (3^(1/2)*5^(1/2))/5 - 3
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 2) - (3^(1/2)*5^(1/2))/5 - 3
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 3) - (3^(1/2)*5^(1/2))/5 - 3
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 4) - (3^(1/2)*5^(1/2))/5 - 3

k =

 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 1)^2/20000 - 3/100000
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 2)^2/20000 - 3/100000
 root(z1^4 + 2*z1^3 - z1^2*((2*3^(1/2)*5^(1/2))/5 + 67/10) - z1*((2*3^(1/2)*5^(1/2))/5 + 36/5) + (36*3^(1/2)*5^(1/2))/25 + 693/50, z1, 3)^2/20000 - 3/100000
 root(z1^4

[Star Strider]

It means there are several solutions. See the documentaion on vpasolve to get numeric solutions and to explore the solution space.

[Duygu Demirpençe]

Ok. i found. Thank you. By the way how can i find iterative approach ?

[Star Strider]

I am not certain what ‘iterative approach’ you are referring to.

Consider using the Optimization Toolbox fsolve function, and a large number of initial parameter estimates to solve for all the roots.