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Department of Science, Math, & Technology

E = mc²
RE: Solve this Quartic equation
Posted: Posted April 9th
Edited April 9th by Xhin
 The Fly posted this a while back: http://gtx0.com/view.php?post=108437 $$(12x-1)(6x-1)(4x-1)(3x-1) = 7$$. I think I've found the secret of this, but my math knowledge isn't great so I have no idea where to go from here. I first looked at the set [12,6,4,3] and tried to find cases where c+d was a multiple of c*d. Both [12,4] and [6,3] are candidates, so I reordered the above expression as: $$(12x-1)(4x-1) = a$$ $$(6x-1)(3x-1) = b$$ $$ab = 7$$. I then factored out both A and B and pulled out those common factors I found before: $$16x(3x-1)+1 = a$$ $$9x(2x-1)+1 = b$$ This is interesting, because it means both a and b are forms of the expression: $$n^2x[(n-1)x-1]$$ a where n = 4 b where n = 3 Unfortunately I have no idea where to go from here. Can someone more math-savvy do something with that?
There are 19 Replies
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It has two solutions. I don’t know what they are, but one is between -1/9 and -1/10, and the other is between 1/2 and 6/11.
I rather doubt they are rational.
They’re probably of the form a*sqrt(7)+b for rational a and b.

Does that help?

I could be wrong, I suppose.

——————————

EDIT: There’s a solution between 25/48 and 19/36.

Edited April 9th by chiarizio

Huh, yeah I didn't realize x could be negative, but you're right.

Posted April 9th by Xhin
Xhin
Nature is beautiful

The positive solution is between 301/576 and 113/216, I think.

And I didn’t try to be math savvy. I just used the “Numbers” spreadsheet program I got with my iPad.

Edited April 9th by chiarizio

I wrote a program to find a fraction that satisfies the equation, and I ran out of precision before I could find one.. the closest is

588744059231737 / 1125899906842624

Posted April 9th by Xhin
Xhin
Nature is beautiful

There's clearly roots or complex fractions happening here.

Posted April 9th by Xhin
Xhin
Nature is beautiful

x = sqrt(9620625609515/35184372088832) is a pretty good approximation. Again, not helpful.

Posted April 9th by Xhin
Xhin
Nature is beautiful

I wound up multiplying it all out so that I could move the 7 over and then graph the thing, and came up with x = -0.1062, x = 0.5229. Simply put, this thing's a hell of a lot more complex than I thought.

Posted April 9th by Black Yoshi

This isn't helpful either

Posted April 9th by Xhin
Xhin
Nature is beautiful

Try solving
(y-1)(y-2)(y-3)(y-4)=168
and then let x=y/12.

Posted April 9th by chiarizio

Working with quadratic equations, once you factor them out you can set each term equal to zero and solve. So, for example, 2x^2 + 5x + 3 = (2x+3)(x+1); setting each of those terms equal to zero gives:

2x+3 = 0, x = -3/2
x+1 = 0, x = -1

I tried that here, but setting each term equal to 7 instead, and came up with:

(12x-1) = 7, x = 2/3
(6x-1) = 7, x = 4/3
(4x-1) = 7, x = 2
(3x-1) = 7, x = 8/3

However, none of these values come close to the solutions I found by graphing the thing. Not to mention that by graphing it I found two solutions, but solving it this way I found 4.

Posted April 9th by Black Yoshi

While poking around, I found this by accident.

Edited April 9th by Xhin
Xhin
Nature is beautiful

I got
Y Y-1 Y-2 Y-3 Y-4 Product X 12x-1 6x-1 4x-1 3x-1 Product
-1.275 -2.275 -3.275 -4.275 -5.275 168.016250390625 -0.10625 -2.275 -1.6375 -1.425 -1.31875 7.00067709960937
-1.274 -2.274 -3.274 -4.274 -5.274 167.820023941776 -0.106166666666667 -2.274 -1.637 -1.42466666666667 -1.3185 6.992500997574
6.274 5.274 4.274 3.274 2.274 167.820023941776 0.522833333333333 5.274 2.137 1.09133333333333 0.5685 6.99250099757399
6.275 5.275 4.275 3.275 2.275 168.016250390625 0.522916666666667 5.275 2.1375 1.09166666666667 0.56875 7.00067709960936

I guess I don’t know how to copy Number spreadsheets into gtx0.

Anyway, the problem for y has a solution between -1.275 and -1.274, and another between 6.274 and 6.275.
Dividing those values by twelve, I get
For x = (-1.275)/12, the product is 7.00067709960937
For x = (-1.274)/12, the product is 6.992500997574
For x = (6.274)/12, the product is 6.99250099757399
For x = (6.275)/12, the product is 7.00067709960936

So those are your solutions within a twelfth of a thousandth.
That’s not a closed-form solution-in-radicals. But using numerical methods I think I got pretty close.

Edited April 9th by chiarizio

Try solving (y-1)(y-2)(y-3)(y-4)=168 and then let x=y/12.

You could try solving (z+1.5)(z+0.5)(z-0.5)(z-1.5)=168 and then let y=z+2.5

Or solve (w+3)(w+1)(w-1)(w-3)=2688 and let z=w/2 so y = (w/2)+2.5 so x = ((w/2)+2.5)/12.

Anyway note we have ((w^2)-9)((w^2)-1) = 2688, which is quadratic in w^2.

Letting v = w^2 this is
(v-9)(v-1) = 2688, or v^2 - 10v + 9 = 2688, or v^2 - 10v - 2679 = 0.

You can now solve for v using the quadratic formula.

v = (10 +- sqrt(10816))/2
So v = (10 +- 104)/2
so v can be either 57 or -47.
w is the square root of v. If v is negative then w is pure imaginary, and so is z = w/2; so y = z+2.5 is complex, and so is x = y/12.

if w is real it is +- sqrt(57)
If w is positive, it is sqrt(57), so z is sqrt(57)/2, so y is (sqrt(57)+5)/2, so x is (sqrt(57)+5)/24.

Posted April 9th by chiarizio

@Black Yoshi: What does the graph look like?

Posted April 9th by Xhin
Xhin
Nature is beautiful

It checks out.
0.522909768136281 5.27491721763537 2.13745860881769 1.09163907254512 0.568729304408844 7

Posted April 9th by chiarizio

@Black Yoshi: What does the graph look like

Like a big curvy W.

by graphing it I found two solutions, but solving it this way I found 4

There are four solutions. A positive real solution; a negative real solution; and two conjugate complex solutions.

Edited April 9th by chiarizio

@chiarizio: I fixed your table, click "edit" to see what I did.

Also that does seem to be breaking the post atm...

Posted April 9th by Xhin
Xhin
Nature is beautiful

If w is positive, it is sqrt(57), so z is sqrt(57)/2, so y is (sqrt(57)+5)/2, so x is (sqrt(57)+5)/24.

You seem to have solved it.

Posted April 9th by Xhin
Xhin
Nature is beautiful

@chiarizio: I fixed your table, click "edit" to see what I did.

So that’s how it’s done! Thanks!

Also that does seem to be breaking the post atm...

I could post the y thing and then post the x thing after it.
And/or I could reduce the precision of the really long decimals.

But do you think I really need to try now?

(5 +- sqrt(57))/24 and (5 +- sqrt(-47))/24.

(Unless I fucked up calculating the complex pair in my head.)

——————————

You can tell they wanted you to solve it because when you do the sqrt(b^2 - 4ac) thing in the quadratic formula, it turns out that 100+4*2679 is a perfect square. If this had been a problem from real life it would probably have been a five-digit prime number or something even uglier.

Edited April 9th by chiarizio
Reply to: RE: Solve this Quartic equation
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