# Why does this method for solving cubics seem to fail?

I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.

OK, so let $$y=ax^3+bx^2+cx+d,$$ where $$a,b,c,d$$ are real and $$a\ne 0.$$ Then $$y'=3ax^2+2bx+c.$$ Now the plan is to consider those cubics as defined above for which $$y'$$ has the same sign for all $$x.$$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.

Now, for $$y'>0,$$ without loss of generality, we must have $$(2b)^2-4(3a)c\lt 0,$$ or $$3ac-b^2>0.$$ Thus, with $$k^2=3ac-b^2,$$ we have $$y'=3a\left[\left(x+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right],$$ by attempting to complete squares. It then follows, since $$y'\ne 0$$ for all $$x,$$ that $$x=\int{\frac{\mathrm d y}{3a\left[\left(y+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right]}}$$ exists for all $$y.$$ This integral can be easily evaluated to give $$\frac1k \arctan{\left(\frac{3ay}{k}+\frac{b}{k}\right)}+C.$$ To determine the constant $$C,$$ note that with the original equation, we have $$y=d$$ when $$x=0,$$ so that we have $$C=-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)},$$ so that $$x=\frac1k \arctan{\left(\frac{3ay}{k}+\frac{b}{k}\right)}-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)}.$$

It is now an easy matter to find the real root $$x$$ of the original equation, which is the value of $$x$$ when $$y=0.$$ This gives the number $$x=\frac1k \arctan{\left(\frac{b}{k}\right)}-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)}.$$

Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $$3^2-3(2)(2)<0,$$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $$x=-3/2.$$ However, applying the method above, we obtain $$k=\sqrt 3,$$ so that the root as given by the arctangent should be $$x=\frac {1}{\sqrt 3} \arctan{\left(\frac{3}{\sqrt 3}\right)}-\frac{1}{\sqrt 3} \arctan{\left(\frac{21}{\sqrt 3}\right)}.$$ If this is so, then we must have $$\frac{1}{\sqrt 3} \arctan{\left(\frac{3}{\sqrt 3}\right)}-\frac{1}{\sqrt 3} \arctan{\left(\frac{21}{\sqrt 3}\right)}=-\frac 3 2,$$ which gives $$\arctan{(7\sqrt 3)}-\arctan{\sqrt 3}=\frac{3\sqrt 3}{2},$$ which is clearly false.

I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!

PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.

• Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks. – NoChance Jul 7 at 23:19
• Sorry if I'm missing something obvious: how did you obtain that $$x=\int{\frac{\mathrm d y}{3a\left[\left(y+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right]}}?$$ – YiFan Jul 7 at 23:33
• I guess this would be from the definition of the inverse function x=f(y). – NoChance Jul 7 at 23:39
• Error. It should be $k^2 =3ac-b^2$. – steven gregory Jul 8 at 0:19
• @stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it. – Allawonder Jul 8 at 0:22

You actually have \eqalign{ & y' = 3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,2} }}} \right)\quad \Rightarrow \cr & \Rightarrow \quad dx = {{dy} \over {3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,2} }}} \right)}} \cr} where you shall read the $$x$$ at the denominator as $$x(y)$$.
• My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it. – Allawonder Jul 8 at 0:15
• @Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless. – Ted Shifrin Jul 8 at 0:27
• @Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ ! – G Cab Jul 8 at 0:36