Problem with solving a quadratic equation

Hello everybody!:)

I would need some help in solving a quadratic equation, where I have to find the two solutions for the variable "z" in function of many parameters

the equation is pretty long and I did the following steps:

syms d N m h b c A U R P k y z
I = ((d+1)/d)*z^(1/d)*N^(1/d)*b-3*c*z^2-U*A*h^k+P*h+R*A-m-h*P^(1+y);    (I is the long euqation)
solve(I)

; the answer I got was:

ans =

log(-(m - P*h + 3*c*z^2 - A*R + A*U*h^k - (N^(1/d)*b*z^(1/d)*(d + 1))/d)/h)/log(P) - 1

Now, are the two solutions coincident? (I do not understand why I only got one of them)

But, main issue, how come I have the variable (z) in the root???? It should not be there!

Maybe I did some mistake..I don't know:/ Any help would be really appreciated!

I apologize if maybe this looks stupid but I am a beginner with matlab

Thank you!

Kodi

Answer 1

[Walter Roberson]

You did not specify which variable to solve for, so it picked "y".

Note: because of the z^(1/d) term, the expression is not quadratic in z.

Answer 2

[Kodi]

Hi Walter, thank you for your kind answer!

Yes, it is not properly quadratic (it's of second degree).

Could you please tell me how can I specify to solve it for z?

Thank you very much!

[Walter Roberson]

solve(I,z)

There is no closed form solution in general, but there might be closed form solutions for particular values of "d"

[Kodi]

True. But..how can I set the equation in matlab, then?

That's what I can't understand:(

[Kodi]

because when I type "Solve(I,z)" it gives me as answer [empty sym] :/

[Walter Roberson]

You can do the solve(). It will return a form involving RootOf(), which the symbolic toolbox knows how to reason about. RootOf(f(x),x) means "the set of values, x, such that f(x) is 0".

When you eventually substitute in enough actual values for your parameters, the Symbolic Toolbox is sometimes able to break down the expression into closed form solutions. When all parameters have been given values, the Symbolic Toolbox is able to find numeric solutions, if you use vpa() or double(). However, the Symbolic Toolbox is often unable to find all solutions in such situations; in some cases it is not able to find any numeric solutions at all even though real-valued solutions exist.

Answer 3

[Kodi]

Thank you very much!

Only, I think I'm doing some typing mistakes as I can't obtain the roots..

If I do not bother you, could you please write me all the passages? (I think it's very quick)

Thank you so much!

[Walter Roberson]

therootssym = solve(((d+1)/d)*z^(1/d)*N^(1/d)*b-3*c*z^2-U*A*h^k+P*h+R*A-m-h*P^(1+y), z);

As noted above, you will likely get a RootOf() answer. After that, if you have values for the variables,

some_values_for_vars = num2cell(rand(1,12));  %whatever values are suitable

theroots_numeric = double( subs(theroomssym, {d, N, m, h, b, c, A, U, R, P, k, y}, some_values_for_vars) )

[Kodi]

Dear Dr. Roberson,

thank you very much for your kind help.

I tried to do this, but when I type

therootssym = solve(((d+1)/d)*z^(1/d)*N^(1/d)*b-3*c*z^2-U*A*h^k+P*h+R*A-m-h*P^(1+y), z);

I subsequently get:

Warning: Explicit solution could not be found. > In solve at 83

and then I'm blocked

:/

[Walter Roberson]

Then you will need to work numerically, which is not unexpected. Use fsolve or something like that.

Define values for all of your variables except z. Then

initial_guess = 1;   %why not
fun = @(z) ((d+1)/d)*z^(1/d)*N^(1/d)*b-3*c*z^2-U*A*h^k+P*h+R*A-m-h*P^(1+y);
approximate_z = fsolve(fun, initial_guess);

[Kodi]

Thank you very much!!