Get Quick Math Help: Solving x^3 + (2x10^5)x^2 + 250x - (5.08x10^-3) = 0 for x

  • Thread starter Thread starter muaddib7
  • Start date Start date
  • Tags Tags
    Homework
AI Thread Summary
To solve the cubic equation x^3 + (2x10^5)x^2 + 250x - (5.08x10^-3) = 0, standard methods like factoring and the quadratic formula are ineffective. Since the equation is cubic, a specific formula for cubic equations should be utilized, which dates back to the 14th century. It's suggested to use numerical methods, such as the bisection method, to find the roots, as they are likely not rational. The complexity of analytic solutions can be overwhelming, making numerical approaches more practical. Ultimately, employing a calculator for numerical methods is recommended for finding the real root.
muaddib7
Messages
1
Reaction score
0
Need to solve the following equation for x

x^3 + (2x10^5)x^2 + 250x - (5.08x10^-3) = 0

I've tried factoring, and quickly found the quadratic formula to be a dead end. Answer has to be a real number. I just need to know how to do it, not necessarily the answer. Thanks all!
 
Physics news on Phys.org
muaddib7 said:
Need to solve the following equation for x

x^3 + (2x10^5)x^2 + 250x - (5.08x10^-3) = 0

I've tried factoring, and quickly found the quadratic formula to be a dead end.
Inasmuch as your equation is a cubic (third-degree polynomial), the quadratic formula doesn't apply. It can be used only on quadratic (second-degree) equations. IOW, equations of the form ax2 + bx + c = 0
muaddib7 said:
Answer has to be a real number. I just need to know how to do it, not necessarily the answer. Thanks all!

There is a formula that can be used to solve cubic equations, that was developed sometime in the 14th century or so by an Italian mathematician. I don't remember the name of the algorithm, but if you searched Wikipedia you might be able to find it.
 
Use a numerical method to find the root, please? The root is very likely not rational and the analytic method will paralyze you with complications, all grace to Cardano. Try bisection first. You can do it on a calculator.
 
Last edited:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top