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# Department of Science, Math, & Technology

E = mc²
Solve this quartic equation
Posted: Posted January 19th, 2018 by The Fly
 Solve the degree 4 equation: $$(12x-1)(6x-1)(4x-1)(3x-1) = 7$$. There's a trick about it that I haven't figured out that helps to solve it, as opposed to solving it directly by multiplying things out. It's a problem from this year's Math Wrangle competition exam. Have fun! The Fly
There are 9 Replies

Never mind the trick, how on earth do you solve this the normal way? I got 0.522909768136281 by writing a program to find the boundaries on each digit.

Posted January 19th, 2018 by Xhin
Xhin
Nature is beautiful

Incidentally I think the trick has something to do with fractions..

Replace 7 with N.

What we want is N=7, but...

If X = 1/3, 1/4, 1/6, 1/12: N=0

If X = 1/2, N=5

If X = 8/12, N = 35

Actually that gives me an idea...

Posted January 19th, 2018 by Xhin
Xhin
Nature is beautiful

Good try! How about rewriting the equation as two quadratic equations? (We can solve each of these directly.)

Posted January 19th, 2018 by The Fly

So, this thing is symmetric and beautiful, but I have no idea where to go from here...

Posted January 20th, 2018 by Xhin
Xhin
Nature is beautiful

Good try! How about rewriting the equation as two quadratic equations? (We can solve each of these directly.)

Yeah but then you have this long expression (including an 864X4) and I can't remember how to solve those.

Edited January 20th, 2018 by Xhin
Xhin
Nature is beautiful

Factor by grouping?

Posted January 20th, 2018 by Black Yoshi

Yes, I know, Xhin, it's not easy even as a degree 4 equation. I believe however that there are methods for dealing with them. I'm familiar with degree 3, but degree 4 have a method which I haven't looked up. In any case, for the quartic equation above it's better not to multiply it all out but instead write it as two solvable quadratic equations like this:

$$(6x?1)(4x?1) = a$$

$$(12x?1)(3x?1) = b$$

where you have $$ab=7$$. This might make the problem more wieldy.

Black Yoshi, I don't think factoring by grouping works in this case since the roots turn out to involve radicals and complex numbers , but you can try it.

Posted January 20th, 2018 by The Fly

I tried taking the log of both sides, and I also tried substitution. So far none of my tricks have helped, and I haven't been able to think of anything else clever. Did you eventually learn the trick?

Posted January 23rd, 2018 by EN

Aside from the quadratic method I mentioned, I'm not aware of any elegant trick. The one I mentioned did work out nicely, though the solutions involve two complex and two real solutions (and not very pretty/easy numbers).

Posted January 27th, 2018 by The Fly
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