Cardano formula

1. Dec 19, 2013

bobie

1. The problem statement, all variables and given/known data
Can we apply the formula to the equation

x^3 -px +q = 0 ?

If not what is the best way to find x ?

2. Dec 19, 2013

SteamKing

Staff Emeritus
I don't see why not.

3. Dec 19, 2013

bobie

I've been trying for days to solve this to no avail, can you show me where I go wrong, please?

.9³ - 400*.9 = 359.271 (359../2 = 179.6355)

³√ (√(179..²+ 400³/27) - 179..) = 11.1075117

³√ (-√(179..²+ 400³/27) - 179..) = -12.0388862 + 11.1075117 = .896

I have tried to change plus to minus around to no avail
Thanks, Steam king

4. Dec 19, 2013

SteamKing

Staff Emeritus
You mistake me for a clairvoyant. Please post the original equation you are trying to solve. Your previous post makes no sense to me.

5. Dec 19, 2013

bobie

x³ -400 x + 359.271 = 0
x= 0.9
applying cardano (q/2 = 179.6355)
± √(q/2² + p³/27 ) -q/2
I get x= 0.896....

Last edited: Dec 19, 2013
6. Dec 19, 2013

Staff: Mentor

It looks like one q/2 needs a sign change, the 400 should be -400, and the resulting square root of a negative leads to complex numbers. (You'll then take into account that each complex number has 3 cube roots.)

I expect that eventually you'll have a handful of candidates, one of which will be the 0.9 you are looking for.

Good luck!

Last edited: Dec 19, 2013
7. Dec 19, 2013

HallsofIvy

Staff Emeritus
If you mean that Cardano's formula says that the root is "± √(q/2² + p³/27 ) -q/2", you are wrong. That isn't Cardano's formula! You have p and q switched and have forgotten a cube root.

If a and b are any two real numbers then
$(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$ and
$-3ab(a+ b)= -3a^3b- 3ab^2$

so $(a+ b)^3- 3ab(a+ b)= a^3+ b^3$

If we let x= a+ b, m= 3ab, and $n= a^3+ b^3$ then $x^3- mx= n$.

Now, suppose we know m and n. Can we find a and b and so x?

Yes. From m= 3ab, b= m/3a so $a^3+ b^3= a^3+ m^3/3^3a^3= n$. Multiply through by $a^3$ to get
$(a^3)^2- (m/3)^3= n(a^3)[itex] or [itex](a^3)^2- n(a^3)- (m/3)^3= 0$, a quadratic equation in $a^3$.

$$a^3= \frac{n\pm\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$

Since $a^3+ b^3= n$, $b^3= n- a^3$ so
$$b^3= \frac{n\mp\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$

Take the cube roots to find a and b, then x= a+ b.

8. Dec 19, 2013

bobie

that is what I did in post 3:

³√ (-√(179.271²+ 400³/27) - 179.271) = -12.00294772 +
³√ (√(179.271²+ 400³/27) - 179.271) = +11.10838241 = 0.89..

Last edited: Dec 19, 2013
9. Dec 19, 2013

bobie

I have tried all sign changes including
-x^3 + 400 x - 359.271 =0
but no 0.9 as a result
There are all real roots, wolfram gives no imaginary root
http://www.wolframalpha.com/input/?i=x^3-400x+359.271=0

Last edited: Dec 19, 2013
10. Dec 19, 2013

Staff: Mentor

11. Dec 19, 2013

Staff: Mentor

And in post #6 I pointed out your mistake, that 400 should be -400.

Wolframalpha is not telling you the whole story. You need to break the evaluation of t1/3 + u1/3 into 2 parts, carefully writing down the 3 roots for t1/3 and then for u1/3. Then perform the final addition/s manually. Out of the 6 answers, you will see one is 0.9

Last edited: Dec 19, 2013
12. Dec 19, 2013

Ray Vickson

Note that $(0.9)^3 - 400 \, (0.9) = -539.271$ (just by striaght calculation), so x = 0.9 is a solution of the equation $x^3 - 400 x + 539.271 = 0.$

13. Dec 19, 2013

scurty

I'm noticing that in all of bobie's posts, he is working with the number 359.271... That might be where his error is.

Edit: Nevermind, the 5 and 3 were mixed up in your post. Sorry.

14. Dec 19, 2013

Ray Vickson

Yes, you are right: it should have been 359.271.

15. Dec 20, 2013

bobie

I tried all sorts of equation with -p
x³ - 33x - 18 =0
18/2 = 9, 33/3 = 11
the solutions are:
http://m.wolframalpha.com/input/?i=solve+z^3+-+33z+-18+=0&x=8&y=7
$$x = 6 ; -5.4495 ; 0.55 05$$
$$a^3 = \frac{18}{2}\pm\sqrt{\left(\frac{18}{2}\right)^2+ \left(\frac{33}{3}\right)^3} → 9 + \sqrt{81+1331} = 46.5766$$$$a = \sqrt[3]{46.5766} = 3.597956362$$
$$b^3 = 9 - \sqrt{81+ 1331} = - 28.5766$$$$b= \sqrt[3]{-28.5766} = -3.05729111$$$$x = a + b = 3.5979 -3.05729 = 0. 54 06$$
changing 9 to -9 gives the same result with inverted sign - 0.54 06
where do I go wrong, please?
or the formula cannot be applied when p < 0 ?

Last edited: Dec 20, 2013
16. Dec 20, 2013

ehild

It is easier to apply some iterative method.

Try xk+1=(xk3+359.271)/400

ehild

17. Dec 20, 2013

bobie

probably to get a negative p we should start from
$(a- b)^3= a^3- 3a^2b- 3ab^2+ b^3$
and two such numbers do not exist

18. Dec 20, 2013

Staff: Mentor

You are making exactly the same mistake as I pointed out yesterday. The Cardano formula should contain (-33/3)^3 but for some reason you keep messing up the sign of the term to be cubed.

Maybe you have written down the formula incorrectly? A readable treatment can be found here: http://www.math.cornell.edu/~dwh/courses/M122-S00/supplements/cardano.html

I can confirm that when you apply the formula correctly, the Cardano method gives the root you are hoping for, viz., 6
Here's a good start: http://m.wolframalpha.com/input/?i=solve+a%3D-33%3B+b%3D-18%3B+t%3D-b%2F2+%2B+sqrt%28%28b%2F2%29%5E2+%2B+%28a%2F3%29%5E3%29%3B+w%3Dt%5E1%2F3&x=0&y=0

Last edited: Dec 20, 2013
19. Dec 21, 2013

bobie

I found b, is that correct?
$$a^3 = \frac{18}{2}\pm\sqrt{\left(\frac{18}{2}\right)^2- \left(\frac{33}{3}\right)^3} â†’ 9 + \sqrt{81-1331} = 46.5766$$$$a = 3+\sqrt{2} i$$
$$b^3 = 9 - \sqrt{81- 1331} = 3- \sqrt{2} i$$
I am not familiar with complex numbers but I suppose that the opposite âˆš2 cancel out
$$a + b = 3+3 + (+ âˆš2i+ - âˆš2i) =0 = 6$$
Is that correct?
but I found b through wolfram, how can I find it by myself? is it too difficult?

Thank you all for your kind help

20. Dec 21, 2013

SteamKing

Staff Emeritus
The roots of polynomials with integer coefficients can be complex. In such cases where there are complex roots, the roots will occur in pairs and be conjugates of one another, i.e. a + bi and a - bi, where a, b are real numbers.