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$$x^4 -2x^3-6x^2-2x+1=0$$ $\left[\text{Hint let } v = x + \frac{1}{x}\right]$

I am stumped, and have no idea how to proceed. I have tried solving it, but have had no success.
 

P.S: This question is meant to be solved, using only techniques for solving quadratic equations.

Mark Viola
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Tobi Alafin
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    Do you mean "find the roots" or something like that? Have you searched for rational roots? Note: there is an obvious rational root. – lulu Jul 22 '17 at 15:56
  • See https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x – lab bhattacharjee Jul 22 '17 at 15:57
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    Ok, now there is a question. In which case my hint is valid, and there is no need for the substitution. – lulu Jul 22 '17 at 15:59
  • Doesn't "solve" mean "find the roots"? A difference in language? – Tobi Alafin Jul 22 '17 at 16:38
  • A simplistic way to find the roots is to graph the polynomial, look to see if there are any obvious rational roots, then divide the polynomial by the factors corresponding to rational roots ($x+1$ in this case). Repeat as many times as you can. – Χpẘ Jul 22 '17 at 19:07

5 Answers5

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I think you meant for $$x^4 -2x^3-6x^2-2x+1=0$$

Obviously $x=0$ is not a solution. So we divide by $x^2$ and get

$$x^2 -2x-6-2x^{-1}+x^{-2}=0$$ $$(x^2+2+x^{-2}) - 2(x+x^{-1}) -8=0$$ $$v^2-2v-8=0$$ $$(v-4)(v+2)=0$$

Jay Zha
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since $x=0$ is not a solution we can divide by $x^2$ $$x^2+\frac{1}{x^2}-2\left(x+\frac{1}{x}\right)-6=0$$ and no set $$t=x+\frac{1}{x}$$ then you will get $$t^2=x^2+\frac{1}{x^2}+2$$ and $$t^2-2=x^2+\frac{1}{x^2}$$ and you have to solve $$t^2-2t-8=0$$

Dr. Sonnhard Graubner
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I would not bother with the substitution. The only possible rational roots are $\pm 1$ and it is easy to check that $-1$ works. A quick calculation shows that your polynomial is $$(x + 1)^2 \,(x^2 - 4 x + 1)$$ and we are done.

lulu
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  • Aren't solutions generally defined on $\mathbb{R}$, and not on $\mathbb{Q}$? – Tobi Alafin Jul 22 '17 at 17:29
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    Not following. I imagined you were looking for complex solutions, no? In any case, searching for rational solutions is a very easy and natural first step...as that solves the problem entirely (well, up to a trivial quadratic) then I don't see any point in looking for a more elaborate solution. – lulu Jul 22 '17 at 17:35
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    To be clear: there was no a priori reason why searching for rational roots should have solved the problem. But then, there was no a priori reason why this particular substitution should have solved it. For what it's worth: In general, there is a closed formula for the roots of quartics (granted, it isn't very nice). – lulu Jul 22 '17 at 17:42
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$x^4 -2x^3-6x^2-2x+1=0$

Divide all terms by $x^2$

$x^2-2x-6-\dfrac{2}{x}+\dfrac{1}{x^2}=0$

Rearrange in this way

$\left(x^2+\dfrac{1}{x^2}\right)-2\left(x+\dfrac{1}{x}\right)-6=0$

Now set $v=x+\dfrac{1}{x}$

squaring you get

$v^2=x^2+\dfrac{1}{x^2}+2\to x^2+\dfrac{1}{x^2}=v^2-2$

Plug in the equation

$v^2-2 -2v -6=0\to v^2-2v-8=0\to v_1=4;\;v_2=-2$

As we want to solve for $x$ two more steps

For $v=4\to x+\dfrac{1}{x}=4 \to x^2-4x+1=0 \to x= 2\pm\sqrt{3}$

for $v=-2\to x+\dfrac{1}{x}=-2\to x^2+2x+1\to x=-1$

Raffaele
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Sometimes you can add and subtract the right thing and land in Pascal's triangle. The 4th row of Pascal's triangle is $1 4 6 4 1$, so add and subtract $6x^2+12x^2+6x$ to get

$$x^4+4x^3+6x^2+4x+1 - 6x^3-12x^2-6x = 0$$

or

$$(x+1)^4 - 6x(x+1)^2=0.$$

Factor out $(x+1)^2$ to get

$$(x+1)^2((x+1)^2 - 6x) = 0$$

or $$(x+1)^2(x^2-4x+1)=0.$$

B. Goddard
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