# stuck with a cubic equation

I picked an equation $0= x^3 +x^2 -2x -1$ I plotted it with geogebra, to see if it had more than $1$ real root. It definitely cuts the $x$-axis $3$ times.

But when I checked wolfram alpha, to see if its roots could be written exactly(I'm doing A-level maths at school, and I have some C3 coursework to do, and if the value of the root can be found by factorising or something I can't use that equation), it said all three roots were complex (here's a link)

I'm just really confused, I can see that it cuts the $x$-axis $3$ times, but it has imaginary parts in all of its roots.

(although it shows the roots can be written exactly, they're complex enough that I can use the equation)

• That's just an artefact of the Wolfram Alpha CAS not correctly doing roots. This is a known problem. If you hit "Approximate Forms", you will get that each one is actually real. Also note that the $y$ scaling on the root plot is $10^{-16}$ so the imaginary parts are very small (particularly, zero). Jan 25 '15 at 19:43
• @MarkMcClure I should clarify and say that Wolfram Alpha (and Mathematica) do not always properly reduce operations on complex numbers. Not that they do not properly do roots, just don't reduce properly and give rather cumbersome expressions. Jan 25 '15 at 19:56
• You have bumped into the Casus Irreducibilis of the cubic, noted first by Bombelli. The Cardano formula "works," sort of, but in using it one cannot avoid roots of non-real quantities. Jan 25 '15 at 19:59

I tried, however, to enter just the equation into Wolfram Alpha without writing solve in front. Then a button appeared saying [Exact forms]. I clicked that button and chose the exact form of $x\approx -1.8$. I copied the exact expression $$x=-\frac13-\frac{7^{2/3}(1+i\sqrt3)}{3\cdot2^{2/3}\sqrt[3]{1+3i\sqrt3}}-\frac16(1-i\sqrt3)\sqrt[3]{\frac72(1+3i\sqrt3)}$$ into a different instance of Wolfram Alpha and asked it to simplify. A bit down the list of results, I found an alternate form reading $$x=-\frac13-\sqrt\frac73\sin\left(\frac13\tan^{-1}(3\sqrt3)\right)-\frac13\sqrt7\cos\left(\frac13\tan^{-1}(3\sqrt3)\right)$$ which surely looks real enough.