Solving x^3 + (2×10^5)x^2 + 250x - (5.08×10^-3) = 0

[muaddib7]

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!

 

[slider142]

This unfortunately doesn't yield well to checking rational roots. In that case, I would go ahead and use one of the formulas for finding the roots of a general cubic.

 

[Mark44]

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

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.

 

[Dick]

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.