# Thread: Solve 1 + 8000x^3=28800x.

1. ## Solve 1 + 8000x^3=28800x.

1 + 8000x^3=28800x

pls am stuck here, how do i get a result for x form the equation above. thanks

2. ## Re: pls need help here (simplify)!

Apparently, it has three roots, but there are no easy formulas for them.

3. ## Re: pls need help here (simplify)!

Set the whole equation to one side and equals zero then factor out this third-degree polynomial.

4. ## Re: pls need help here (simplify)!

Originally Posted by theloser
Set the whole equation to one side and equals zero then factor out this third-degree polynomial.
... that I would like to see.

5. ## Re: pls need help here (simplify)!

Maybe you can do something with Cardano's formula but that's also complicated so maybe a math program is useful to solve this one.

6. ## Re: pls need help here (simplify)!

KISS ... use a calculator.

7. ## Re: Solve 1 + 8000x^3=28800x.

Originally Posted by lawochekel
1 + 8000x^3=28800x

pls am stuck here, how do i get a result for x form the equation above. thanks
You need to post the exact instructions you were given for this question.

8. ## Re: Solve 1 + 8000x^3=28800x.

There are various numerical methods to find a solution. Once you pin one down, the other two are easy. That first on is the tricky one, but, off the top of my head, it can't be very far from zero (0). You tell me why this is so.

Shoot for x = 0.000034722222234 and see how far it gets you. Tell me how YOU managed to find it.

9. ## Re: Solve 1 + 8000x^3=28800x.

3 imaginary solutions:

Online Equation Solver

Guess you could "pretend" you never saw the "1"; then:
1600x(x^2 - 18) = 0

10. ## Re: Solve 1 + 8000x^3=28800x.

Not a bad idea! The "1" is so small compared to the other coefficients that this should give you a reasonable "first approximation". Then use a simple "perturbation method" to improve the approximation. If your first order approximation is $\displaystyle x_0$, let $\displaystyle x= x_0+ y$ to get a cubic equation in y. Make the same kind of reduction to get an approximate value, $\displaystyle y_0$ for y, then repeat with $\displaystyle x= x_0+ y_0+ z$.