- #1

bobie

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## Homework Statement

Can we apply the formula to the equation

x^3

**-p**x +q = 0 ?

If not what is the best way to find x ?

Thanks for your help

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- Thread starter bobie
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- #1

bobie

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Can we apply the formula to the equation

x^3

If not what is the best way to find x ?

Thanks for your help

- #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?

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

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

I have tried to change plus to minus around to no avail

Thanks, Steam king

- #4

SteamKing

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- #5

bobie

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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.

Good luck!

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!

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- #7

HallsofIvy

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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]

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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that is what I did in post 3:take the cube roots to find a and b, then x= a+ b.

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

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

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- #9

bobie

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I have tried all sign changes includingIt looks like one q/2 needs asign 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!

-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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- #10

NascentOxygen

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If you don't get the signs right, you can't hope to arrive at the solution you are expecting.

- #11

NascentOxygen

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

Wolframalpha is not telling you the whole story. You need to break the evaluation of t

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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

- #13

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Note that ##(0.9)^3 - 400 \, (0.9) = -539.271## (just by striaght calculation), so x = 0.9isa 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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I tried all sorts of equation withThis equation and no other?

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 - 0.5

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 x

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 band so x?

.

probably to get a negative

[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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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.where do I go wrong, please?

or the formula cannot be applied when p < 0 ?

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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[tex]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 [/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 âˆš2 cancel out

[tex] a + b = 3+3 + (+ âˆš2i+ - âˆš2i) =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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- #21

NascentOxygen

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Looks better. (I can't see how "= 46.5766" comes into the picture, though.)

[tex]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 [/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 âˆš2 cancel out

[tex] a + b = 3+3 + (+ âˆš2i+ - âˆš2i) =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

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. 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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Thanks for your kind help,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]

I have noticed that it is not difficult, the square root of a negative âˆš-n is just the root of the positive

plus i, if âˆšn =a, [itex]\sqrt[3]{n}[/itex]= b , then âˆš-n = ai, and [itex]\sqrt[3]{-n}[/itex] just

b/2 [itex]\pm[/itex] b/2 * iâˆš3.

Isn't there a hard-and-fast rule to get the cube root of m + âˆš-n?

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- #23

bobie

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Thanks for the advice.It is easier to apply some iterative method.

Try x_{k+1}=(x_{k}^{3}+359.271)/400

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

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!

http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]

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