# Solutions for roots polynomials.

1. Jul 30, 2006

### MathematicalPhysicist

i wonder if there's a formula like for quadartic and cubic equations also for roots polynomials, like this equation:
ax^(1/2)+bx+c=0
or like
ax^(1/3)+bx^(1/2)+cx+d=0
?
and what are they?

Last edited: Jul 30, 2006
2. Jul 30, 2006

### 0rthodontist

y = x^(1/2) turns the first into a quadratic equation in y, and y = x^(1/6) turns the second into a polynomial equation in y of degree 6.

3. Jul 30, 2006

### Data

strictly speaking a polynomial in $x$ must be a linear combination of natural powers of $x$ (so your equations aren't quite polynomial equations). 0rthodontist's suggestion will convert your equations to polynomial equations, though. You'll just have to reject some of the solutions ($x^{1/3}$ and $x^{1/2}$ are 1-1 once you choose a given branch, while $x^2$ and $x^6$ are many-1). Of course, solving degree 6 polynomial equations by hand isn't exactly fun either (unless the coefficients are cooperative, the best way is usually to try to approximate the roots with something like Newton's method, because there's no closed-form solution for the roots of polynomials of order $\ge 5$).

Last edited: Jul 30, 2006