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Having a bit of a problem with a question. There is a 4m ladder leaving against a wall. There is a box in between The ladder and wall. The box is a cubic metre.

I have found a quartic to find the length up the wall the ladder will reach. But its proving difficult for me to solve.

$$x^4 +2x^3 -14x^2 +2x +1=0$$

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    $\begingroup$ The downvotes are piling up. And they're right: You haven't even asked a question, just stated that you find some equation difficult to solve. Is it the equation you want help with, or do you wonder if you have the right equation? I suppose you are aware that there is a formula for solving the general quartic, though it rather big and complicated. (And this one clearly has no rational roots.) $\endgroup$ – Harald Hanche-Olsen Oct 20 '14 at 12:55
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    $\begingroup$ The standard method for a quartic in this form is to divide through by $x^2$ and use $y=x+\frac 1x$ to reduce to solving a quadratic for $y$ and then a second quadratic for $x$. $\endgroup$ – Mark Bennet Oct 20 '14 at 12:55
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    $\begingroup$ "In this form" refers to the fact that the sequence of coefficients of the polynomial is symmetric with respect to its midpoint, here (a,b,c,b,a). $\endgroup$ – Did Oct 20 '14 at 13:01
  • $\begingroup$ @MarkBennet I had forgotten that one. I guess it's been a few decades since I had to solve a quartic. $\endgroup$ – Harald Hanche-Olsen Oct 20 '14 at 13:04
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    $\begingroup$ Another feature of the equation is that $\frac 1y$ is a solution whenever $y$ is a solution. I haven't checked, but it looks as though $x$ is the height up the wall excluding the height of the box. Then you expect a reciprocal equation, because the distance along the ground is also a solution, and the reciprocal property is given by similar triangles. $\endgroup$ – Mark Bennet Oct 20 '14 at 14:55
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You can find more about this type of equations under the name reciprocal equation or reciprocal polynomial.

See, for example, also this post Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ (And several of the posts shown there among linked questions.)


In this particular case, you have: $$ \begin{align} x^4+2x^3-14x^2+2x+1&=0\\ x^2+2x-14+\frac2x+\frac1{x^2}&=0 \end{align} $$

If we use the substitution $u=x+\frac1x$, we get $u^2=x^2+2+\frac1{x^2}$ $\Rightarrow$ $x^2+\frac1{x^2}=u^2-2$. So your equation gets to the form $$ \begin{align} \left(x^2+\frac1{x^2}\right)+2\left(x+\frac1x\right)-14&=0\\ u^2+2u-16=0\\ (u+1)^2-17=0 \end{align} $$ which yields $$u_{1,2}=-1\pm\sqrt{17}.$$

Now we have to solve for each $u$ the equation $$ \begin{align} x+\frac1x=u\\ x^2-ux+1=0 \end{align} $$ which yields $$x=\frac{u\pm\sqrt{u^2-4}}2.$$ Since we have $u^2=16-2u$, this can be rewritten as $$x=\frac{u\pm\sqrt{12-2u}}2.$$ So the solutions are $$ \begin{align} x=\frac{-1\pm\sqrt{17}+\sqrt{14\mp2\sqrt{17}}}2 \end{align} $$

Here is what WolframAlpha returns for your equation: (Link) Note that you can switch there between approximate and exact forms of the result.

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  • $\begingroup$ BTW is there in English also a different name for this kind of equations than reciprocal equations? $\endgroup$ – Martin Sleziak Oct 20 '14 at 14:47
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    $\begingroup$ Commenting on your comment. Polynomials of degree $n$ such that $x^nP(1/x)=P(x)$ are often called palindromes or palindromic. I don't know how official it is. The term is kinda self-explanatory. $\endgroup$ – Jyrki Lahtonen Sep 3 '16 at 21:13
  • $\begingroup$ @JyrkiLahtonen Official enough to be mentioned on Wikipedia. I did not know this terminology, thanks for letting me know. $\endgroup$ – Martin Sleziak Sep 3 '16 at 21:18

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