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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
SteamKing said:I don't see why not.
that is what I did in post 3:hallsofivy said:take the cube roots to find a and b, then x= a+ b.
I have tried all sign changes includingNascentOxygen said: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!
And in post #6 I pointed out your mistake, that 400 should be -400.bobie said: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..
bobie said: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
Ray Vickson said: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.##
scurty said: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.
I tried all sorts of equation with -pNascentOxygen said:This equation and no other?
bobie said: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
HallsofIvy said: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?
.
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.bobie said:where do I go wrong, please?
or the formula cannot be applied when p < 0 ?
Looks better. (I can't see how "= 46.5766" comes into the picture, though.)bobie said: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
Thanks for your kind help,NascentOxygen said: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
Thanks for the advice.ehild said:It is easier to apply some iterative method.
Try xk+1=(xk3+359.271)/400
ehild said: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
Cardano's formula is a method for solving cubic equations, which are equations in the form ax^3 + bx^2 + cx + d = 0. It was developed by Italian mathematician Gerolamo Cardano in the 16th century and involves finding the roots of the equation.
To apply Cardano's formula, you first need to rewrite the equation in a specific form. Then, you use the formula to calculate three values, called the roots, which are the solutions to the equation. These roots can be real or complex numbers.
The steps to solving a cubic equation using Cardano's formula are: 1) Rewrite the equation in the form x^3 + px + q = 0, where p and q are constants. 2) Calculate the value of a constant called the discriminant. 3) Use the discriminant to determine the type of roots the equation has. 4) Use Cardano's formula to calculate the three roots of the equation. 5) Check your answers by plugging the roots back into the original equation.
Yes, there are limitations to using Cardano's formula. It can only be used to solve cubic equations, and it is not always possible to express the roots in a simple form. Additionally, the calculations involved can be complex and time-consuming.
Cardano's formula has applications in fields such as engineering, physics, and economics. It can be used to find solutions to problems involving cubic equations, such as finding the maximum or minimum value of a function. It is also used in computer graphics and animation to create smooth curves and surfaces.