# Does the limit of the cubic formula approach the quadratic one as the cubic coefficient goes to $0$?

The formula for solving a cubic equation of the form $$ax^3+bx^2+cx+d=0$$ does not seem to yield the quadratic formula for the limit $$\lim _{a \rightarrow 0} \text{(cubic formula)}$$.

But, if one tries the same thing with the quadratic formula the limit exists for the right choice of the square root sign.

My question is, is there a way to take the limit $$\lim _{a \rightarrow 0} \text{(cubic formula)}$$ and produce the quadratic formula? or does the limit simply not exist? Finally, if the limit does not exist, is there a technical reason for that?

Any input is very much appreciated. Thanks

Edit:

I was trying to tackle the simple case of $$b=0$$. Doing so, the cubic formula reduces to

$$\left(-\frac{d}{2a} - \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}} + \left(-\frac{d}{2a} + \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}$$

If one only considers the first term cubed, then it can be written as

$$\frac{1}{a}\left(-\frac{d}{2} - \left(4c^3+27ad^2\right)^{\frac{1}{2}}\right)$$

and I don't see how is it possible to find a finite limit as $$a\rightarrow 0$$. Am I missing something trivial?

• Well it is going to be complicated - I would start by working with $y=3ax$ to clear the fractions, so there are no divisions by $a$. Also note that with the cubic formula being the sum of two cube roots, you have three choices for each - nine in total, but only three of these choices give roots of the original equation. Then you probably need to pick the right one of these. (This is the equivalent of picking the right sign for the square root). You can test for compatibility by taking the product of the two cube root terms, which is fixed. May 10 at 12:40
• Limit exists, since there are many different ways to get roots in radicals, including for $a=0$ returning the solution of the quadratic equation. May 10 at 13:00
• Although it is overkill in this specific setting, I think you can proceed by Rouche's theorem: consider a disk around $x_0$ which is a zero of the quadratic, picking it in a range so that on the boundary you have $|ax|^3<|bx^2+cx+d|$, but (if $c^2-4bd \neq 0$) not so large that you surround the other zero. In the nondegenerate case you can pick $r$ to be $2 \frac{|ax_0|^3}{|2bx_0+c|}$ for sufficiently small $a$. The degenerate case is actually easier because you just have to avoid the large root of the cubic.
– Ian
May 10 at 13:11
• The analogous question for one degree less is: Does the limit of the quadratic formula $x \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ approach the linear one $x = \frac{-c}{b}$ as the quadratic coefficient goes to 0? Does it?
– Dan
May 10 at 22:10
• @Dan Yes, write it as $\,x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}\,$ and ignore the root that goes to infinity. Or use the same approach as in my answer, and take the reciprocal equation $\,c\, (1/x)^2 + b\,(1/x) + a\,$ with $\,a\to 0\,$.
– dxiv
May 10 at 22:26

When $$a \to 0$$ one of the roots goes to infinity, which complicates the algebraic manipulations. Instead, it is easier to show that when $$d=0$$ the non-zero roots reduce to the quadratic formula.

Assume WLOG $$\,a=1\,$$, then with the wikipedia notations for the cubic formula in the case $$d=0\,$$:

\begin{align} \Delta_{0} = b^{2}-3ac &= b^2 - 3c \\ \Delta_{1} = 2b^{3}-9abc+27a^{2}d &= 2b^3 - 9bc \\ C = {\sqrt[{3}]{\frac{1}{2}\left(\Delta _{1}\pm {\sqrt {\Delta _{1}^{2}-4\Delta _{0}^{3}}}\right)}} &= \sqrt[3]{\frac{1}{2}\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right)} \tag{1} \end{align}

The radicals in $$(1)$$ can be denested, since it is straightforward to verify that:

$$\left(-b \pm \sqrt{-3(b^2-4c)}\right)^3 = 4\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right) \tag{†}$$

It follows that:

\require{cancel} \begin{align} C &= \frac{1}{2}\left(-b \pm \sqrt{-3(b^2-4c)}\right) \tag{2} \\ \frac{\Delta_0}{C} = \frac{2(b^2-3c)}{-b \pm \sqrt{-3(b^2-4c)}} &= \frac{1}{2}\left(-b \mp \sqrt{-3(b^2-4c)}\right) \tag{3} \end{align}

Finally, the roots are $$\displaystyle\, x = -{\frac {1}{3}}\left(b + \omega C+{\frac {\Delta_0}{\omega C}}\right)\,$$ where $$\,\omega^3 = 1\,$$, and $$\,\omega = 1\,$$ gives the root $$\,x=0\,$$. Otherwise, with $$\,\omega = \dfrac{-1 \pm i\sqrt{3}}{2}\,$$ a complex cube root of unity, using $$(2)$$ and $$(3)\,$$: \begin{align} x &= -\frac{1}{3}\left(b + \omega\,\frac{-b \pm \sqrt{-3(b^2-4c)}}{2} + \overline\omega\,\frac{-b \mp \sqrt{-3(b^2-4c)}}{2}\right) \\ &= -\frac{1}{3}\left(b \cdot \left(1-\frac{\omega + \overline\omega}{2}\right) \pm \sqrt{-3(b^2-4c)} \cdot \frac{\omega - \overline\omega}{2} \right) \\ &= -\frac{1}{3}\left(\frac{3}{2}\, b \pm \frac{i \sqrt{3}}{2} \sqrt{-3(b^2-4c)} \right) \\ &= \frac{-b \pm \sqrt{b^2-4c}}{2} \end{align}

The latter matches the quadratic formula, as expected.

[ EDIT ] $$\;$$ The denesting of $$\,C = {\sqrt[{3}]{\frac{\Delta _{1}}{2}\pm {\sqrt {\frac{\Delta _{1}^{2}}{4} - \Delta _{0}^{3}}}}}\,$$ in $$\,(\dagger)\,$$ follows from my answer here.

a sufficient condition for $$\,\sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$$ to denest is for $$\,m^2 \cdot p - n^2 \cdot q\,$$ to be the cube of a rational $$\,r\,$$, and for the cubic $$\,p\, t'^{\,3} - 3r\, t' - 2m\,$$ to have an appropriate rational root

Replacing "rational" with "rational expressions in the coefficients", the sufficient conditions are satisfied in this case, since:

• $$\,m^2 \cdot p - n^2 \cdot q = \Delta_0^3\,$$ is a perfect cube;

• $$\,t'^{\,3} - 3 \Delta_0 t' - \Delta_1 = t'^{\,3} - 3(b^2-3c) t' - b(2b^2-9c)\,$$ has the root $$\,t'=-b\,$$.

• So, for the case $d=0$, you would expect such a behavior since you can factor out an $x$ from the cubic formula (maybe not ?). I am not sure that this answers my question. Is it not feasible to take the limit $a \rightarrow 0$ since one of the roots always diverges? Or is there a way to isolate the "bad" root? Thanks for the detailed explanation when $d=0$.
– Joel
May 11 at 8:03
• @Joel That's precisely the expected behavior - the cubic formula gives one root as $0$ and the other two reduce to the quadratic formula. This does fully answer your question, it's just a matter of relabeling the coefficients of the reciprocal equation. Say your original cubic is $dx^3 +cx^2+bx+a=0$ with $d \to 0$. Then make the substitution $z=1/x$ and you get the equivalent cubic $az^3+bz^2+cz+d=0$. The above shows that the roots reduce to the quadratic formula when $d \to 0$, in fact when $d=0$ since you no longer need to take limits.
– dxiv
May 11 at 8:22
• I see, this clears things up. Thanks for the detailed answer.
– Joel
May 11 at 13:46