Solve for real $x$


By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe that there must be some nice method to solve this which I can't think about. Please help. Thanks!

  • $\begingroup$ Another root is $-\varphi$, where $\varphi$ is the golden ratio. The third belongs to the so-called irreducible case. $\endgroup$ – Lucian Jun 2 '14 at 7:01
  • $\begingroup$ Are you allowed to use inspection ? As you said, once you arrived to the quintic, you are stuck ! $\endgroup$ – Claude Leibovici Jun 2 '14 at 8:10

You shouldn't dismiss Karo's graph. Drawing a graph to get a feel for the equation (when you don't know how to proceed) is most helpful. And this isn't nonsense, in fact the graph is the key for the solution, as it makes it clear that we can consider only $x$ for which $-2 \leq x \leq 2$. This can be conveniently rewritten as $-1\leq\frac{x}{2}\leq1$.

This inequality reminds us immediately of the one that $\sin a$ and $\cos a$ also satisfy, and indeed we can assume WLOG $\frac{x}{2}=\sin a$, for some angle $a$ in radians of course. This makes $x=2 \sin a$, so let's substitute this in the equation.

$$\begin{align} 8\sin^3 a-6\sin a&=\sqrt{2(\sin a+1)}\\ -2(-4\sin^3a+3\sin a)&=\sqrt{2(\sin a+1)}\\ -2\sin3a&=\sqrt{2(\sin a+1)} \end{align}$$

It helped that we could factor it into $\sin 3a$, but then again we can't get rid of the square root in a nice way. This is because $\sin^2a$ is given in terms of $\cos 2a$, and not $\sin 2a$, so we can't turn the radicand into a nice square of a sine. On the other hand, $\cos^2 a$ can be expressed in terms of a cosine (indeed, we have the trigonometric identity $\cos^2a=\frac{1+\cos2a}{2}$). So we are lead to believe that the substituition $x=2\cos a$ is much more fortunate. It's worth a shot:

$$\begin{align} 8\cos^3 a-6\cos a&=\sqrt{2(\cos a+1)}\\ 2(4\cos^3a-3\cos a)&=\sqrt{4\frac{(\cos a+1)}{2}}\\ 2\cos3a&=\sqrt{2^2\cos^2\frac{a}{2}}\\ \cos3a &=\cos\frac{a}{2} \end{align}$$

The equation has been successfully trivialized. We should note that imposing $0\leq a\leq\pi$, $\cos a$ will still assume all of the values it possibly could, so we impose this restriction on $a$.

Therefore, we can have $3a=\frac{a}{2}\Rightarrow a=0$, which gives us the solution $x=2\cos0=2$, or we can have the other, non-trivial solutions: $$3a=\frac{a}{2}+2\pi\Rightarrow a=\frac{4\pi}{5}$$ which gives us $x=2\cos\frac{4\pi}{5}=-2\cos\frac{\pi}{5}$. If $0\leq a\leq\pi$, then $0\leq 3a \leq 3\pi\lt4\pi$, so we don't need to look the case $3a=\frac{a}{2}+4\pi$ or higher, and we just need to consider the last case: $$3a=2\pi-\frac{a}{2}\Rightarrow a=\frac{4\pi}{7}$$ Which gives us the last solution, $x=2\cos\frac{4\pi}{7}=-2\cos\frac{3\pi}{7}$.

Note: This solution was inspired by the one brazilian mathematician Nicolau C. Saldanha presented in a lesson. Here's the link: http://y2u.be/jFFdOSsVGgg (someone who's fluent in Spanish shouldn't have trouble understanding the explanations). It is also shown that with some simple manipulations on the regular pentagon, one can arrive at $\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$, so the second solution can be written as $x=-\frac{1+\sqrt{5}}{2}$ (which justifies the quadratic factor $x^2-x-1$ of the 6th degree polynomial you get when you square both sides, which obviously contains some extraneous roots).

I'm not sure there's an easy way to turn $\cos\frac{3\pi}{7}$ into one algebraic term (if there is one at all), so we just leave it that way. The solution set is $$x \in \left. \left \{ 2,-\frac{1+\sqrt{5}}{2},-2\cos\frac{3\pi}{7} \right. \right \}$$

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    $\begingroup$ One could also finish the solution using the sine substituition. To get rid of the square root, write $\sin a=\cos(\frac{\pi}{2}-a)$, factor it to get rid of the square root, then transform it back into a sine. But everything goes smooth if you use cosine instead of sine. $\endgroup$ – Deathkamp Drone Jul 31 '14 at 11:09

The graph is like this and have 3 answers not one answer

enter image description here

  • $\begingroup$ I'm sorry but the question is not about finding the number of solutions, but about solving for them. $\endgroup$ – Henry Jun 2 '14 at 7:04
  • $\begingroup$ Alt solutions said that it have only one answer I just post graph that proves that it has 3 answers, I don't have enough reputations to leave comment. $\endgroup$ – Karo Jun 2 '14 at 7:09

By raising both sides to two: $$x^6-6 x^4+9 x^2-x-2 = 0,$$ (and $x^{3}-3x\geq 0$)

By decomposition you get:

$$(x-2) (x^2+x-1) (x^3+x^2-2 x-1) = 0$$

therefore you have 6 roots:





$$x\approx -0.445$$

$$x\approx 1.246$$

But you have the constraint: $x^{3}-3x\geq 0$

So, only three of the solutions are acceptable: $$x=2.0$$

$$x\approx-1.618$$ $$x\approx-0.445$$

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    $\begingroup$ Can you explain the factorization? $\endgroup$ – tpb261 Jun 2 '14 at 7:00
  • $\begingroup$ Actually $x\approx-1.618$ also works. And $x\approx0.445$ should be $-0.445$, which works too. $\endgroup$ – David Jun 2 '14 at 7:01
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    $\begingroup$ A quick glance at the graph give 3 acceptable roots, 2 of which are less then zero. $\endgroup$ – Ron Gordon Jun 2 '14 at 7:02
  • $\begingroup$ @Alt How did you think about the factorization? $\endgroup$ – Henry Jun 2 '14 at 7:02
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    $\begingroup$ @tpb261 The factorisation will need to be of form $(x^2+ a x \pm 1)(x^3+ b x^2+ c x+ \pm 1) = x^5+2x^4-2x^3-4x^2+x+1$, and that's not too hard to solve for integer / rational coeffts, if they exist. $\endgroup$ – Macavity Jun 2 '14 at 8:09

Let $$x+2=y^2$$ Therefore $$\sqrt{x+2}=y$$ for any positive $y$. Rewriting the whole expression in $y$, we get, $$y^6-6y^4+9y^2-y-2$$ which factorizes to $$(y-2)(y^2+y-1)(y^3+y^2-2y-1)$$ You only want the positive roots of the polynomial because $y$ can't be negative. You can easily obtain the roots of the linear, the quadratic and the cubic equation. After you've obtained the values of $y$, get the corresponding values of $x=y^2-2$. No need to solve a quintic polynomial. :)

  • $\begingroup$ how is the degree 6 polynomial factorized to three factors? $\endgroup$ – tpb261 Jun 2 '14 at 7:46
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    $\begingroup$ I'm sorry I don't understand your question. Are you asking me how I managed to factorize the polynomial? I just used a Computer Algebra System called maxima. Or if you're asking how a degree 6 polynomial can be factorized into 3 factors, notice that factor 1 has degree 1, factor 2 has degree 2 and factor 3 has degree 3 so they add up to 6. $\endgroup$ – sayantankhan Jun 2 '14 at 7:51
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    $\begingroup$ I was asking how did you factorize the polynomial. I got the answer: " Computer Algebra System called maxima". $\endgroup$ – tpb261 Jun 2 '14 at 7:53

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