I'm stumped on this question. Is there any way of solving a cubic that doesn't have rational roots?

Question

The question is $x^3 - 20x^2 = 8x^2 - 180$.

What I've tried is moving everything to one side, so, $x^3 - 28x^2 + 180 = 0$. Then, solving this like I would solve a cubic, I tried to find all the rational roots. After using the Rational Root Theorem, I found none.

I tried using Mathway to solve but they just told me to graph it and then find the solutions, which were really long decimals.

Is there any way of like, special factoring or a trick to solve this by hand?

There is a formula, but it is so complicated that people often find numeric approximation a better approach. — Ross Millikan, Feb 28 '23 at 03:37
this isn't nice at all, as you noticed, but if you really wanted a closed form for the roots, you could convert this into a depressed cubic and then use Cardano's method. — Hersh, Feb 28 '23 at 04:05
There's really no nice trick for this particular cubic: the roots simply do not lend themselves to such. In particular this cubic is an example of the so-called casus irreducilibus for the cubic formula. This does, however, mean there is a trigonometric formula for the roots. But this requires Hersh's suggestion about converting to a depressed cubic. — Semiclassical, Feb 28 '23 at 04:10

score 1 · Answer 1 · answered Feb 28 '23 at 07:32

It is true that the solutions of cubic equations can be ugly (have a look here) : however, this is true only if you use Cardano method.

Follow the steps given here : you have $$\Delta=14930640 \qquad\qquad p=-\frac{784}{3}\qquad\qquad q=-\frac{39044}{27}$$ So, three real roots. So, use the trigonometric solution to get $$x_k=\frac{28}{3} \left(1+2 \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{9761}{10976}\right)\right)\right)\right)$$ with $k=0,1,2$.

Is it so ugly ?

I'm stumped on this question. Is there any way of solving a cubic that doesn't have rational roots?

1 Answers1