BUG - Cardano's solution returns wrong answer

[miladsade96]

For a_1 = 1, a_2 = 2, a_3 = 3 ,
it should print
x_1 = -1.27568
z_2 = 0.13784 + i * 1.52731
z_3 = 0.13784 - i * 1.52731
But it prints
x_1 = -2.58565688
z_2 = ( 0.792838216 , -2.33903098 )
z_3 = ( 0.792719483 , 2.33903098 )

I think the error might be at sign section of code:
https://github.com/EverLookNeverSee/FCS/blob/acb60b0728226dfd4508fa420787b606294a74ba/src/numerical%20methods/Exercise_14.f90#L36

[miladsade96]

@LKedward Do you have any idea about this?

[LKedward]

Hi @everlookneversee , I've been through the code and run it on my computer and I can't find any mistake with respect to the formulae you are referencing .
(As you point out, the code still doesn't return the correct result with sign(1.0,R).)

i.e. I believe you have implemented the formulae correctly, so I wonder whether there's a mistake in the formulae?
Unfortunately I can't verify the formulae since I can't find a reference online that formulates Cardano's method in the same way as here - I would recommend either verifying the formulae from another source or working off a different set of formulae.

[miladsade96]

Hi @everlookneversee , I've been through the code and run it on my computer and I can't find any mistake with respect to the formulae you are referencing .
(As you point out, the code still doesn't return the correct result with sign(1.0,R).)

i.e. I believe you have implemented the formulae correctly, so I wonder whether there's a mistake in the formulae?
Unfortunately I can't verify the formulae since I can't find a reference online that formulates Cardano's method in the same way as here - I would recommend either verifying the formulae from another source or working off a different set of formulae.

Thank you very much

[miladsade96]

@LKedward
I found solution below and would like to use it:

[miladsade96]

Solved!