Quartic equation?

A

apple2357

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Mar 9, 2018
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I found this problem online ( from a Finnish website) but can't find a solution, I am stuck now, having only found one value by inspection? Which i can share but i dont want to lead.
I feel there must be a simpler way of doing this than the obvious, i.e. generating a quartic etc.

Solve

8/(x-8)+10/(x-6)+12/(x-4)+14/(x-2)=4

The symmetry in the values suggests something but I can't see anything else. Any other suggestions? No need to solve it. Just ideas would be great
 
what if we let [MATH]u=x-5[/MATH][MATH] \dfrac{4}{u-3}+\dfrac{5}{u-1} + \dfrac{6}{u+1} + \dfrac{7}{u+3} = 2 [/MATH]
Doing this and simplifying everything one solution immediately pops out.
Another one is available by a bit of detailed inspection. (Think rational root theorem)

Then you can divide the original quartic (in [MATH]u[/MATH]) by the quadratic these two roots produce.

That leaves a quadratic that can be easily solved with the quadratic formula.

Note of course you'll have to add [MATH]5[/MATH] to each value of [MATH]u[/MATH] to get the corresponding value of [MATH]x[/MATH].
 
Romsek said:
what if we let [MATH]u=x-5[/MATH][MATH] \dfrac{4}{u-3}+\dfrac{5}{u-1} + \dfrac{6}{u+1} + \dfrac{7}{u+3} = 2 [/MATH]
Doing this and simplifying everything one solution immediately pops out.
Another one is available by a bit of detailed inspection. (Think rational root theorem)

Then you can divide the original quartic (in [MATH]u[/MATH]) by the quadratic these two roots produce.

That leaves a quadratic that can be easily solved with the quadratic formula.

Note of course you'll have to add [MATH]5[/MATH] to each value of [MATH]u[/MATH] to get the corresponding value of [MATH]x[/MATH].
Click to expand...

Thats a neat idea. I got x=16 by inspection and recognising 1+1+1+1 =4
and then thought aout the 2,4,6,8 in the denominators and felt there must be other things.
 
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