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.
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.
Huh, yeah I didn't realize x could be negative, but you're right.
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.
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
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
There's clearly roots or complex fractions happening here.
There's clearly roots or complex fractions happening here.
x = sqrt(9620625609515/35184372088832) is a pretty good approximation. Again, not helpful.
x = sqrt(9620625609515/35184372088832) is a pretty good approximation. Again, not helpful.
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.
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
https://i.imgur.com/GZuWQgl.png
This isn't helpful either
Try solving
(y1)(y2)(y3)(y4)=168
and then let x=y/12.
Try solving
(y1)(y2)(y3)(y4)=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:
(12x1) = 7, x = 2/3
(6x1) = 7, x = 4/3
(4x1) = 7, x = 2
(3x1) = 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.
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:
(12x1) = 7, x = 2/3
(6x1) = 7, x = 4/3
(4x1) = 7, x = 2
(3x1) = 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.
https://i.imgur.com/w1bYFBa.png
While poking around, I found this by accident.
I got
[table top full]
Y  Y1  Y2  Y3  Y4  Product  X  12x1  6x1  4x1  3x1  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
[/table]
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 closedform solutioninradicals. But using numerical methods I think I got pretty close.
I got
Y  Y1  Y2  Y3  Y4  Product  X  12x1  6x1  4x1  3x1  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 closedform solutioninradicals. But using numerical methods I think I got pretty close.
Edited April 9th
by chiarizio
> Try solving (y1)(y2)(y3)(y4)=168 and then let x=y/12.
You could try solving (z+1.5)(z+0.5)(z0.5)(z1.5)=168 and then let y=z+2.5
Or solve (w+3)(w+1)(w1)(w3)=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
(v9)(v1) = 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.
Try solving (y1)(y2)(y3)(y4)=168 and then let x=y/12.
You could try solving (z+1.5)(z+0.5)(z0.5)(z1.5)=168 and then let y=z+2.5
Or solve (w+3)(w+1)(w1)(w3)=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
(v9)(v1) = 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?
@Black Yoshi: What does the graph look like?
It checks out.
0.522909768136281 5.27491721763537 2.13745860881769 1.09163907254512 0.568729304408844 7
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.
@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...
@chiarizio: I fixed your table, click "edit" to see what I did.
Also that does seem to be breaking the post atm...
> 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.
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.
> @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 fivedigit prime number or something even uglier.
@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 fivedigit prime number or something even uglier.
Edited April 9th
by chiarizio