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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?

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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?

    The answers are
    (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
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