?>

# 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
Page: settingsSettings

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
Enter your message here
 #newbits { margin-bottom: 10px; } #newbits a, #pop a { background-color: #000; padding: 5px; margin-right: 5px; border: 1px solid #333; text-decoration: none; font-size: 0.8em; box-shadow: 0px 0px 11px #444 inset; position: relative; } #newbits a:hover, #pop a:hover { background-color: #035; } #newbits a b, #pop a b { color: #efefef !important; font-weight: normal !important; font-style: none !important; text-decoration: none !important; } .newbits-back { box-shadow: 0px 0px 11px #700 inset !important; } #pop { display: none; padding: 3px; background-color: #111; box-shadow: 0px 0px 10px #000 inset; margin-bottom: -1px; } #pop a { display: inline-block; padding: 0px 12px; margin-left: 0px; margin-right: 10px; } #pop a.sel { color: #c0ac5d; } #quickr.popped { position: fixed; bottom: 0px; right: 0px; z-index: 10; box-shadow: 0px 0px 24px 15px #000; } #quickr.popped.left { right: initial; left: 0px; } #quickr.popped.top { bottom: initial; top: 0px; } #quickr.popped #pop { display: block; } #pop a i { position: relative; } #quickholder { height: 240px; } Add Image @ Handy Reference Guide: = Makes paragraph bolded // Makes paragraph italic !cyan Makes paragraph cyan !yellow Makes paragraph yellow !red Makes paragraph red !green Makes paragraph green !purple Makes paragraph purple !pink Makes paragraph pink !orange Makes paragraph orange !=cyan Makes paragraph bolded cyan !//cyan Makes paragraph italicized cyan This works for the other colors above ** Gives a line solid circle bullets *- Gives a line solid square bullets *=Gives a line outline circle bullets [b]TEXT[/b] Bolds text [i]TEXT[/i] Italicizes text [s]TEXT[/s] Strikes through text ~(draw)~ and ~(roll)~ and ~(flip)~ work... try them out :P <3 = red-purple heart

 Site Rules | Complaints Process | Give Feedback Facebook Page GTX0 © 2009-2019 Xhin GameTalk © 1999-2008 lives on