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Abbey78336

Member

Registered: 2022-08-15

Posts: 37

Negative second degree equation

Can any one solve this ?

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

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)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

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

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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

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,876

Re: Negative second degree equation

Hi imcute;

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...

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"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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Negative second degree equation

hi

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

This spreadsheet will solve most cubics.

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

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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

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,876

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:

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)

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"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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

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

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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