I have a problem, from Gelfand's "Algebra" textbook, that I've been unable to solve, here it is:

Problem 268.

What is the possible number of solutions of the equation $$ax^6+bx^3+c=0\;?$$

Thanks in advance.

  • $\begingroup$ Do you allow complex solutions, or just real? $\endgroup$ – Ross Millikan Jan 12 '13 at 16:38

Hint: $\quad$Let $y = x^3$:

$$ax^6 + bx^3 + c = 0 \quad \iff \quad ay^2 + by + c = 0\tag{1}$$

Solve for $y$ ... there will be either two real solutions, one real solution, or no real solutions when solving for $y$ (why?, when?). (Examine the discriminant.)

  • When is $\Delta = \sqrt{b^2 - 4ac}\; < \;0\,$? And what does this mean in terms of the existence (or non-existence), of real solutions in y?

  • When $\Delta = 0$, there is exactly one real-valued solution $y$.

  • When $\Delta > 0$, there are two unique real-valued solutions $y_1, y_2$.

In each case, then, for each (possible) solution $y_i$ of the right-hand equation in $(1)$, what are the number of solutions in $x$ to $y_i = x^3$ for each solution $y_i$? (Note that the degree $3$ is odd in $y = x^3$, so we don't have to worry whether solutions ($y$'s) are positive or negative. If $y_i$ is a solution, then there will exist $x$ such that $y = x^3$.) Simply check cases for each possible root $y_i$.

  • $\begingroup$ Nice illustration! You really pictured the problem. +1 $\endgroup$ – mrs Jan 15 '13 at 16:34
  • $\begingroup$ Thank you so much for such detailed explanation, I had no internet connection and couldn't reply for some time. My problem was that I didn't realize that $(x^3)^2=x^6$ haha, yeah I am that stupid, so when I saw your answer, starting with "Let $y=x^3$", I immediately realized what the solution would be. So thank you. $\endgroup$ – Paul Dirac Jan 16 '13 at 10:17

Put $\,t:=x^3\,$ , so your equation becomes

$$(*) at^2+bt+c=0\Longrightarrow \Delta= b^2-4ac$$

Now, if $\,\Delta=0\,\,$ then $\,(*)\,$ has one unique solution. $\,x^3=t={-b/2a}\,$ , and if $\,\Delta >0\,$ then there're two solutions for $\,t=x^3\,$.

Since $\,3\,$ is an odd natural we don't care whether the solutions above are positive or negative, there always are solutions as long as $\,\Delta\geq0\,$, so now you have to take care of the different cases...

  • 1
    $\begingroup$ Simple and encouraging way. +1 $\endgroup$ – mrs Jan 12 '13 at 17:22

Hint: Set $t=x^3$ and solve for $t$


The aplication of quadratic formula in $$ a\cdot (x^3)^2+b\cdot(x^3)+c=0 $$ give us that the possible roots enjoy
$$ x^3 =\left[\frac{-b+\sqrt{b^2+4ac}}{2a}\right] \quad \mbox{ or } \quad x^{3} =\left[\frac{-b-\sqrt{b^2+4ac}}{2a}\right] $$ I believe that the greatest difficulty is to extract all the cubic roots of these expressions. For every cubic roots to use de Moivre's formula. Extracting the cubics roots of $\left[\frac{-b+\sqrt{b^2+4ac}}{2a}+ i\cdot 0\right]$ and $\left[\frac{-b-\sqrt{b^2+4ac}}{2a} + i\cdot 0\right]$ by de Moivre's formula we have this equation have six roots $$ x_{+\,i} =\sqrt[3\;]{\frac{-b+\sqrt{b^2+4ac}}{2a}}\cdot \omega^i \quad \mbox{ and } \quad x_{-\,i} =\sqrt[3\;]{\frac{-b-\sqrt{b^2+4ac}}{2a}}\cdot \omega^i $$ where $i=0,1,2$ and $\omega$ is a complex cubic root of unit, for exemple $$ \omega=\frac{1}{2}+i\cdot\frac{\sqrt{3}}{2} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.