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#1 2022-10-01 19:00:45

Abbey78336
Member
Registered: 2022-08-15
Posts: 37

Negative second degree equation

Can any one solve this ?


2x^(-2) + 5x + 10 = 0

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#2 2022-10-12 12:33:18

imcute
Member
Registered: 2022-09-28
Posts: 176

Re: Negative second degree equation

first you get rid of fractions
5x^3+12=0
then take the cube root
x^3=-12/5
x=pow(-12/5,1/3)
done!
ps:my method does not involve quadratic formulas so there may be some lost roots
edit:sorry!wrogn math!
first you get rid of fractions
5x^3+10x^2=-2
then you factor out 5x^2
x^2(x+2)=2/5
then you search google for third degree formulas and get desperate and call for help
@bob ~

Last edited by imcute (2022-10-12 12:37:49)

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#3 2022-10-13 10:40:40

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Negative second degree equation

hi imcute

I've been working on a spreadsheet calculation for this.  Just completed but the answer has come out wrong so I've got to de-bug it.

The formulas are pretty complicated so it's not surprising and it's getting late here in the UK so I think I'll re-start in the morning.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2022-10-13 21:42:10

phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,810

Re: Negative second degree equation

Hi imcute;

imcute wrote:

5x^3+10x^2=-2
then you factor out 5x^2
x^2(x+2)=2/5

The resulting equation after factoring out 5x² is missing something...


"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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#5 2022-10-14 23:44:16

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Negative second degree equation

hi

Finally got the de-bug completed (I hope).

This spreadsheet will solve most cubics.

gpod0WU.gif

The original cubic values are input at line 3 and are labelled above as a, b, c, and d.

The sheet works in three stages.

Stage one. I am using anonimnystefy's first step here: https://www.mathisfunforum.com/viewtopic.php?id=17660

The substituation creates a new cubic with a zero y^2 term.  Line 5 shows the formulas to create the new cubic. Line 6 shows the labels e, f, g, and h for the new cubic.   f is, of course, 0 but I have included it as I think it helps to keep the coefficients aligned.

I have not used his next stage but rather zetafunc's given here:

https://www.mathisfunforum.com/viewtopic.php?id=27921 post 9.

The two methods are similar; I stuck with zetafunc's as I'm used to using it.

Stage two. Create a quadratic from the cubic. Line 9 shows the quadratic in z; line 10 the formulas to make it; and line 11 the calculated values.

Stage three. The positive root solution is determined and from this q, then p, then p+q = y, and finally x values determined by working the substitutions backwards.

I have tested with a number of cubics. Sometimes a coefficient comes out as zero and leads to a divide by zero error but these cases can usually be solved more simply anyway. The one in the screen shot is the cubic from this thread.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2022-10-17 22:49:05

phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,810

Re: Negative second degree equation

Hi;

I thought I'd try Excel's Goal Seek, and managed to solve this thread's cubic and a couple of others I tested it on:
UeUq0W6.png

The two extra cubics have 3 roots each, and to find them I entered into column D the initial x values that I'd predetermined and stored in Column C, those values being:
(a) a high initial value to try to find the highest x;
(b) a low initial value to try to find the lowest x;
(c) half the sum of the resulting highest & lowest x values from (a) & (b) to find the remaining x (not sure if that's a good idea generally, but it worked for these equations).

Then I ran Goal Seek (individually for the equations in rows 2, 3, 5-7 & 10-12) with the respective Goal Seek boxes filled from columns E & D as per the image example.

Btw, to increase Goal Seek's precision (quite a necessary step for good accuracy with this method), I first set 'Maximum Change' in Excel's File/Options/Formulas/'Enable iterative calculation' to 0.000000000000001, which gave accuracy to 14 decimal places for this thread's cubic: i.e., x ≈ -2.09144638072228 (the last few digits aren't displayed in the image).

The above cubics are the only ones I've tried this Goal Seek method on...

Last edited by phrontister (2022-10-18 10:29:10)


"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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#7 2022-10-18 22:52:24

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Negative second degree equation

hi Phro,

Thanks for that.  I have tended to avoid goal seek because of the limited precision. Silly me; didn't know you can improve it so many thanks for that.  There are several other problems that I can apply that to.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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