MHB My TOP Favorite Polynomial Challenge

[anemone]

Like I mentioned in the title, this is probably one of the greatest challenge problems (I've seen so far) that designed for, hmm, well, for a challenge!

Let $x_1$ be the largest solution to the equation

$\dfrac{6}{x-6}+ \dfrac{8}{x-8}+\dfrac{20}{x-20}+\dfrac{22}{x-22}=x^2-14x-4$

Find the exact value of $x_1$.

 

[kaliprasad]

great question

Spoiler

add 1 to each term onLHS and so 4 to RHS to get

$\dfrac{x}{x-6} + \dfrac{x}{x-8} + \dfrac{x}{x-20} + \dfrac{x}{x-22} = x^2-14x$

so one solution is x = 0 and further we deviding by x we get

$\dfrac{1}{x-6} + \dfrac{1}{x-8} + \dfrac{1}{x-20} + \dfrac{1}{x-22} = x-14$

put y = x - 14 to get

$\dfrac{1}{y+8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} + \dfrac{1}{y-8} = y$

or

$\dfrac{1}{y+8} + \dfrac{1}{y-8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} = y$

or

$\dfrac{2y}{y^2-64} + \dfrac{2y}{y^2-36} = y$

so y = 0

or

$\dfrac{2}{y^2-64} + \dfrac{2}{y^2-36} = 1$

or

$2((y^2-36) + y^2-64))= (y^2-36)(y^2-64)$

or $2((2y^2-100))= (y^2-36)(y^2-64)= y^4-100y^2+ 36 *64$

or $y^4- 104y^2+48^2+200=0$

or(y^2-52)^2 = 200

$y^2 = 52 \pm 10\sqrt{2}$

we should take the higher of the 2 and add 14 to get the largest x or $x = 14 + \sqrt{52+10\sqrt{2}}$ as y = x- 14

 

[anemone]

Well done, kaliprasad!(Yes) Thanks for agreeing with me that this is a great problem() and thanks for participating!