solve quadratic equation problem

Hi guys, I am trying to solve a quadratic equation for Mu, but when i use solve (f,Mu) an X show up which is not in the original equation. i do not know how.

the quadratic equation;

(((V1*K1)/(K1+((4*Mu)/3)))+((V2*K2 )/(K2+((4*Mu)/3))))+ 5*(((V1*Mu2)/(Mu-Mu2))+((V2*Mu1)/(Mu-Mu1)))+2; =0

f=(((V1*K1)/(K1+((4*Mu)/3)))+((V2*K2 )/(K2+((4*Mu)/3))))+ 5*(((V1*Mu2)/(Mu-Mu2))+((V2*Mu1)/(Mu-Mu1)))+2;

solve(f,Mu)

Answer 1

[Walter Roberson]

It isn't a quadratic equation, it is a quartic equation. The explicit solutions for it total over a million characters long.

Did you notice the RootOf that showed up in the answer? See http://www.mathworks.com/help/symbolic/mupad_ref/rootof.html

[Abdulaziz]

yes you alright it is quartic equation. thank you for your replay Walter always appreciate your help.

I saw the link you sent and I am trying to understand how the p,q equations can present the roots of Mu. I found a similar code where uses p,q,r,s equations to find Mu but I do not know what are these equations.

[Walter Roberson]

You probably got something more like Example 3, RootOf(expression_in_z, z) where the roots are the quantities, z, such that expression_in_z becomes 0. These are not explicit values, and (in general) are not even necessarily analytically computable (but they are for quartics): the RootOf() structure represents "whatever works to make the expression 0"

To find out what the roots actually are for the quartic, you would need to ask solve() to expand up to degree 4. The result will be very messy, though, and it is doubtful you will be able to really grasp them intuitively. It gets a bit easier when you substitute actual values in for the variables, but even then and even with pretty simple coefficients, values show up that tend to make me go "Huh??" For example even just x^4+x^3+x^2+x+1 has a solution that involves a term that is a square root of an expression that involves a term that is a square root of an expression that involves a term that is the square root of 375. I have no feel at all for how one gets from the very simple coefficients of all 1's, to multiply-nested square roots of 375... and your expression is much worse.