Finding real roots of a cubic equation

I have the equation:

eqn = ((d^3*pi*((4251*d + 5951400)/(25*d))^(1/2))/(2*(d + 1400)))*(pi*(d^2)/4)-180 == 0

and want to find the real root/roots.

when i attempt to solve it with

x = vpasolve(eqn,d)

Matlab only return the imaginary part of the solution:

- 3.593452147062167996782934136112 - 1.3074862631155484137468498544529i

How do i find the real solution?

Answer 1

[Roger Stafford]

This is an equation that can be manipulated so that d is one of the roots of a ninth degree polynomial equation of which only one of its nine roots is real.

The original equation is:

((d^3*pi*((4251*d+5951400)/(25*d))^(1/2))/(2*(d+1400)))*(pi*(d^2)/4)==180

Since 4251*d+5951400 = 4251*(d+1400), the d+1400 partially cancels with the same quantity in the denominator and we get the equation

pi^2/40*d^5*(4251/(d*(d+1400)))^(1/2)==180

or

pi^2/40*d^5*4251^(1/2) == 180*(d*(d+1400))^(1/2)

Squaring both sides and transposing gives

4251/1600*pi^4*d^10-32400*d^2-45360000*d == 0

One factor d can be factored out since d = 0 is clearly not a solution of the original equation and that finally leaves the polynomial equation

4251/1600*pi^4*d^9-32400*d-45360000 == 0

The ‘roots’ function can be used for this and it shows that there is only one real root, namely

d = 3.82617917662798

Answer 2

[Massimo Zanetti]

To force the vpasolve finding only real solutions you cannot use assume. If you know the real solution is only one a workaround is to set search range to [-Inf,Inf]:

x = vpasolve(eqn,d,[-Inf,Inf])
x =

3.8261791766279723703687735908663

By the way, I suggest you to read documentation to get more insights: vpasolve

[robin johansson]

Thank you for the answer! Just what i was looking for.

With this problem i did know that i only had 1 real solution. I've read the documentation and i still don't understand how i would go about if i didn't know how many real solutions i could find.

My idea of how to get around this would be

X = vpasolve(eqn,x,[-inf, inf],'random',true)

Y = vpasolve(eqn,x,[X, inf],'random',true)

Z = vpasolve(eqn,x,[-inf, X],'random',true)

etc.

Not sure if this works, could you suggest a smarter solution to finding all real values of a nonpolynomial equation?

[Massimo Zanetti]

For non polynomial equations there is no general rule to find all solutions. Consider using solve function. The documentation is rich of useful examples that cover many potential applications.

Answer 3

[Walter Roberson]

One approach for polynomials is to use solve instead of vpasolve, to get all of the solutions, and then to use logical indexing to select the results whose imag()==0

[Walter Roberson]

sols = solve(eqn, x);
sols(imag(sols)~=0) = [];   %remove the ones that have a non-zero imaginary component.

However, in R2016b (and perhaps some other releases), this code can fail due to a bug in the Symbolic Toolbox (I notified them of the bug a few weeks ago.) The work-around for the moment is

sols(imag(vpa(sols))~=0) = [];