The solutions of this equation are given. They are $\frac{\sqrt2}{2}$, $\frac{\sqrt6 -\sqrt2}{4}$ and $-\frac{\sqrt6 +\sqrt2}{4}$. However i'm unable to find them on my own. I believe i must make some form of substitution, but i can't find out what to do. Help would be very appreciated. Thanks in advance.

  • 4
    $\begingroup$ It is natural I think to start by letting $x=\sqrt{2}\,t$. And then one hopes that the resulting equation has a rational root. $\endgroup$ Dec 22, 2015 at 19:20
  • $\begingroup$ Cubic function roots and see especially Cardano's method and Vieta's substitution. $\endgroup$ Dec 22, 2015 at 19:23
  • $\begingroup$ Did you try Ruffini? $\endgroup$
    – Charly
    Dec 22, 2015 at 19:23
  • $\begingroup$ @StephenG yes i know the general formula. But i'm told that there is a method to find it by a proper substitution, possibly a trigonometric substitution i think. I say this because the two roots stem from angles of $π/12$. And yes $\sqrt2 / 2$ is the easy root. The other two are the hard ones. $\endgroup$
    – KeyC0de
    Dec 22, 2015 at 19:24
  • 1
    $\begingroup$ After you did what @AndréNicolas said, you will be able to realize that $t=\frac{1}{2}$ is a root, i.e., $x=\frac{\sqrt2}{2}$ is a root of the equation in question. $\endgroup$
    – frosh
    Dec 22, 2015 at 19:26

3 Answers 3


HINT...Let $x=\cos\theta$

The equation becomes $\cos3\theta=-\frac{1}{\sqrt{2}}$

Can you take it from there?

  • $\begingroup$ Very, very clever. $\endgroup$
    – John
    Dec 22, 2015 at 19:36

Using the substitution $x=\frac{t}{\sqrt{2}}$ the equation becomes $$4{t^3} - 6t + 2 = 0,$$ Using Ruffini you get that $$4{t^3} - 6t + 2 = \left( {t - 1} \right)\left( {4{t^2} + 4t - 2} \right) = 2\left( {t - 1} \right)\left( {2{t^2} + 2t - 1} \right)$$ This hows that one solution is $t=1$, i.e. $x=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$. To obtain the other two solutions just apply the discriminant formula to the equation ${2{t^2} + 2t - 1}=0$.


Let $x=\sqrt{2}\,t$. Substituting, we find that the equation becomes $16\sqrt{2}\,t^3-6\sqrt{2}\,t+\sqrt{2}=0$, which is equivalent to $$16t^3-6t+1=0.$$ Now one hopes that the above equation has a rational root. By the Rational Roots Theorem, the possible candidates are $\pm 1$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm \frac{1}{8}$, and $\pm \frac{1}{16}$.

Note that $t=\frac{1}{2}$ works. Now divide the cubic by $t-\frac{1}{2}$ to look for other roots.

Remark: It is a little nicer to note that $t$ is a root of our cubic if and only if $t=\frac{1}{s}$, where $s$ is a root of $s^3-6s^2+16=0$. Then we have the pleasure of not working with fractions.

  • $\begingroup$ It seems that $x=\frac{t}{\sqrt2}$ substition makes things easier. $\endgroup$
    – frosh
    Dec 22, 2015 at 19:38
  • $\begingroup$ @frosh: Yes it does. I was taking things one step at a time. But after the $t$ substitution, it would have been better to let $2t=w$, which is the same as your suggestion. $\endgroup$ Dec 22, 2015 at 19:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .