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Solve the equation $x^4 - 14x^3 + 50x^2 -14x + 1 = 0$.
I am not sure about how to best proceed, and would like a solution that does not involved the generalised quartic formula.

YDP
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3 Answers3

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Hint: First observe the equation is palindromic. Divide throughout with $x^2$ and rewite it as a quadratic in $\left(x+\dfrac{1}{x}\right)$.

sciona
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    What is the exact form of the polynomial after you rewrite it in your suggested way? I get $x^2 - 14x + 50 - \frac{14}{x} + \frac{1}{x^2}$, but not sure how to proceed – YDP Nov 26 '14 at 10:59
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    next make $x^2+\dfrac{1}{x^2} = \left(x+\dfrac{1}{x}\right)^2 - 2$ .. – sciona Nov 26 '14 at 11:00
  • The exact answers are $x = 3 \pm \sqrt{2}, 4 \pm \sqrt{15}$ – YDP Nov 26 '14 at 21:34
  • @Kinhu: The first one looks like you may have lost a factor of 2 before the sqrt(2)... I've just done it in my head so I may be wrong but might be worth double checking... – Chris Nov 26 '14 at 22:37
  • Yes, sorry, you are correct – YDP Nov 28 '14 at 16:53
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Not knowing the substitution trick, you can anyway infer that if $x$ is a solution, then $1/x$ as well, so that the polynomial can be factored in two polynomials of the second degree, and these will be palindromic too:

$$x^4 - 14x^3 + 50x^2 -14x + 1 =(x^2+Ax+1)(x^2+Bx+1).$$ Developing and identifying, $$A+B=-14,\\1+AB+1=50.$$ The solutions are $$\frac{-14\pm\sqrt{14^2-4\cdot48}}2=-8,-6.$$

Now solve $$x^2-8x+1=0,\\x^2-6x+1=0.$$

  • I notice that with some palindromic quartic polynomials, such as $x^4+2x^3+4x^2+2x+1$, the above technique still works, but $A$ and $B$ become non-real. In fact $x^4+2x^3+4x^2+2x+1 = (x^2+(1+i)x+1)(x^2+(1-i)x+1)$. Of course this can still be used to find the roots, none of which are real. When all roots are known (they are "ugly" in my example), we might pair them up in a different way to get the decomposition into real quadratic polynomials (which will be non-palindromic) whose existence is known from the real-number version of the fundamental theorem of algebra. – Jeppe Stig Nielsen Nov 28 '14 at 16:18
  • @JeppeStigNielsen: yep, you can factor in two real polynomials, but how do you achieve that ? You need to somehow exploit palindromy (actually meaning that there are just two independent coefficients) otherwise you fall back on the general quartic problem. –  Nov 28 '14 at 17:12
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A more detailed solution:

If we divide the equation by $x^2$:

$$\frac{x^4}{x^2} - \frac{14x^3}{x^2} + \frac{50x^2}{x^2} - \frac{14x}{x^2} + \frac{1}{x^2} = x^2 - 14x + 50 - \frac{14}{x} + \frac{1}{x^2}$$

Then, combining like terms, we notice that:

$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50$$

If we let $y = x+\frac{1}{x}$

Note that:

$$\left(x+\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2$$

Therefore,

$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50 = y^2 -2 -14y + 50 =0$$

$$y^2 - 14y +48 =0$$

$$(y-6)(y-8) = 0$$

Therefore, $x + \frac{1}{x} = 6$ and $x + \frac{1}{x} = 8$

Can you take it from here?

Varun Iyer
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