The largest root of $-3x^3+24x^2+6x-9=0$

Question

Since the polynomial has three irrational roots, I don't know how to solve the equation with familiar ways to solve the similar question. Could anyone answer the question?

Answer 1

I used Mathematica to obtain the roots using the cubic formula, which I then simplified with the ComplexExpand function. The root which is largest in absolute value is

$$ x_\max = \frac{8}{3}+\frac{2}{3} \sqrt{70} \cos\left[\frac{1}{3} \arctan\left(\frac{9 \sqrt{2351}}{1087}\right)\right] \approx 8.19930. $$

The other two roots are

$$ x_\pm = \frac{8}{3}-\frac{1}{3} \sqrt{70} \cos\left[\frac{1}{3} \arctan\left(\frac{9 \sqrt{2351}}{1087}\right)\right]\pm\sqrt{\frac{70}{3}} \sin\left[\frac{1}{3} \arctan\left(\frac{9 \sqrt{2351}}{1087}\right)\right], $$

which are approximately

$$ x_+ \approx 0.513388 $$

and

$$ x_- \approx -0.712687. $$

Answer 2

The roots of cubic polynomials are obtained in a way which is similar to what is done with quadratic polynomials. I suggest you look here.
The basic idea is to transform a $x^3 + b x^2 + c x + d =0$ to the so-called depressed cubic which, after a simple change of variable, write $t^3 + p x + q = 0$. Then, the sign of $(4 p^3 + 27 q^2)$ tells the nature of the roots and its value allows to compute these roots whatever the could be (real or complex).