Cardano formula

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  • #1
bobie
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Homework Statement


Can we apply the formula to the equation

x^3 -px +q = 0 ?

If not what is the best way to find x ?

Thanks for your help
 

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  • #2
SteamKing
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I don't see why not.
 
  • #3
bobie
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I don't see why not.

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
SteamKing
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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
bobie
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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....
 
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  • #6
NascentOxygen
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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. :smile:

Good luck!
 
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  • #7
HallsofIvy
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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
[itex](a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex] and
[itex]-3ab(a+ b)= -3a^3b- 3ab^2[/itex]

so [itex](a+ b)^3- 3ab(a+ b)= a^3+ b^3[/itex]

If we let x= a+ b, m= 3ab, and [itex]n= a^3+ b^3[/itex] then [itex]x^3- mx= n[/itex].

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

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

The quadratic formula gives
[tex]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}[/tex]

Since [itex]a^3+ b^3= n[/itex], [itex]b^3= n- a^3[/itex] so
[tex]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}[/tex]

Take the cube roots to find a and b, then x= a+ b.
 
  • #8
bobie
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take the cube roots to find a and b, then x= a+ b.
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..
 
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  • #9
bobie
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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. :smile:Good luck!
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
 
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  • #11
NascentOxygen
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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..
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
 
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  • #12
Ray Vickson
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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

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
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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.##

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
Ray Vickson
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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.

Yes, you are right: it should have been 359.271.
 
  • #15
bobie
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This equation and no other?
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
[tex]x = 6 ; -5.4495 ; 0.55 05[/tex]
[tex]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 [/tex][tex]a = \sqrt[3]{46.5766} = 3.597956362 [/tex]
[tex]b^3 = 9 - \sqrt{81+ 1331} = - 28.5766[/tex][tex]b= \sqrt[3]{-28.5766} = -3.05729111[/tex][tex]x = a + b = 3.5979 -3.05729 = 0. 54 06[/tex]
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 ?
 
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  • #16
ehild
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Homework Statement


Can we apply the formula to the equation

x^3 -px +q = 0 ?

If not what is the best way to find x ?

Thanks for your help

It is easier to apply some iterative method.

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

ehild
 
  • #17
bobie
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If a and b are any two real numbers then
[itex](a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex] and
[itex]-3ab(a+ b)= -3a^3b- 3ab^2[/itex]
.....................
Now, suppose we know m and n. Can we find a and b and so x?
.

probably to get a negative p we should start from
[itex](a- b)^3= a^3- 3a^2b- 3ab^2+ b^3[/itex]
and two such numbers do not exist
 
  • #18
NascentOxygen
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where do I go wrong, please?
or the formula cannot be applied when p < 0 ?
You are making exactly the same mistake as I pointed out yesterday. :cry: 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
 
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  • #19
bobie
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I found b, is that correct?
[tex]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 [/tex][tex]a = 3+\sqrt{2} i [/tex]
[tex]b^3 = 9 - \sqrt{81- 1331} = 3- \sqrt{2} i [/tex]
I am not familiar with complex numbers but I suppose that the opposite √2 cancel out
[tex] a + b = 3+3 + (+ √2i+ - √2i) =0 = 6 [/tex]
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
SteamKing
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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.
 
  • #21
NascentOxygen
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I found b, is that correct?
[tex]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 [/tex][tex]a = 3+\sqrt{2} i [/tex]
[tex]b^3 = 9 - \sqrt{81- 1331} = 3- \sqrt{2} i [/tex]
I am not familiar with complex numbers but I suppose that the opposite √2 cancel out
[tex] a + b = 3+3 + (+ √2i+ - √2i) =0 = 6 [/tex]
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
Looks better. (I can't see how "= 46.5766" comes into the picture, though.)

If you aren't up with complex numbers, then you'll have to be like those in Cardano's day and restrict your use of the method to those cases where the value under the square root evaluates as non-negative.

Wolframalpha makes maths more interesting, it gives you an enticing glimpse of what lies over the next crest. :smile: I don't think you could determine the cube root of a complex number easily manually.
Enjoy your study! http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
 
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  • #22
bobie
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If you aren't up with complex numbers, then you'll have to be like those in Cardano's day and restrict your use of the method....
Enjoy your study! http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
Thanks for your kind help,
I have noticed that it is not difficult, the square root of a negative √-n is just the root of the positive
plus i, if √n =a, [itex]\sqrt[3]{n}[/itex]= b , then √-n = ai, and [itex]\sqrt[3]{-n}[/itex] just
b/2 [itex]\pm[/itex] b/2 * i√3.
Isn't there a hard-and-fast rule to get the cube root of m + √-n?
 
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  • #23
bobie
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It is easier to apply some iterative method.
Try xk+1=(xk3+359.271)/400
Thanks for the advice.
Could you show me how to do that on a pocket calculator or tell me how many operations it requires? it must be surely slower, but how much slower
 
  • #24
ehild
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Start with an easy trial xo value, say, xo=0
Type in the trial value, take the cube.
Add 359.271 (and store 359.271) . Divide the result by 400.
Getting the result (x1) take the cube, add the retrieved value (359.271) divide the result by 400.... and so on.

xo=0

x1=359.271/4=0.898. Substituting back:
x2=0.89999
substituting back again
x3=0.8999999, x4=0.9. It took a few seconds with my stone-age pocket calculator.

If you start with xo=1
x1=0.9006775
x2=0.9000041
x3=0.9

If you have one root, divide the original equation by (x-0.9). That results a quadratic equation.

ehild
 
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  • #25
bobie
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x1=0.9006775
x2=0.9000041
x3=0.9
If you have one root, divide the original equation by (x-0.9). That results a quadratic equation. ehild

Thanks, ehild, that was vere quick!
Merry Christmas to everybody!
:smile:http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
 
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