@Calvin Lin
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Please Calvin,can you teach me CARDANO'S METHOD...........PLEASEEEEEE
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Alpha Beta
·
4 years, 7 months ago

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@Alpha Beta
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For Cardano's method, I'd suggest just reading up on it. It's a bunch of algebra stating how to solve the depressed cubic (no quadratic term).

There is also a general formula for the cubic (and quartic) equation, but I can honestly say that I never bothered to memorize what they are.
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Calvin Lin
Staff
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4 years, 7 months ago

[LaTex edited in by peter...not sure if I preserved Tim's order's of operations correctly (lots of parentheses), so see his top section for official version]
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Tim Ye
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4 years, 7 months ago

I don't mean to be intrusive, but I edited in a LaTex attempt of your answer so that people can read it easier. If that is not what you meant, feel free to re-edit it, using my LaTex and the formatting guide for help.

There is a link to the formatting guide at the base of the window where you enter a comment. Please put all math in LaTex as it is way easier to read than teasing out terms bounded by parentheses within parentheses. LaTex is actually easier, or just as hard as typing math the ugly way, but it is still easy to bungle. Please use the edit post button and the preview function, until your math states what you mean. Cheers:)
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Peter Taylor
Staff
·
4 years, 7 months ago

## Comments

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TopNewestYou must know the product of the roots and the sum of the roots, you can also use that – Gurpreet Singh · 4 years, 6 months ago

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You can use rational root theorem to know if there is a rational root, or you can cyclic factorize it. – Gurpreet Singh · 4 years, 6 months ago

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– Gurpreet Singh · 4 years, 6 months ago

However, there is no rational rootLog in to reply

how do u get this....... – Alpha Beta · 4 years, 7 months ago

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– Alpha Beta · 4 years, 7 months ago

Please Calvin,can you teach me CARDANO'S METHOD...........PLEASEEEEEELog in to reply

There is also a general formula for the cubic (and quartic) equation, but I can honestly say that I never bothered to memorize what they are. – Calvin Lin Staff · 4 years, 7 months ago

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– Alpha Beta · 4 years, 7 months ago

1,2,3,4,5Log in to reply

i want full solution please/........ – Alpha Beta · 4 years, 7 months ago

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hmmm,,, i'd leave the question if i were you

anyways, you can see solutions here

Click on exact form (its dirty) – Harshit Kapur · 4 years, 7 months ago

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x = -1/6 (1-i sqrt(3)) (27/2-(3 sqrt(69))/2)^(1/3)-((1+i sqrt(3)) (1/2 (9+sqrt(69)))^(1/3))/(2 3^(2/3))

\(x = \frac{-1}{6} (1-i \sqrt{3}) (\frac{\frac{27}{2}-(3 \sqrt{69}}{2})^\frac{1}{3}-(1+i \sqrt{3}) \frac{\frac{1}{2} (9+\sqrt{69})^\frac{1}{3}}{(23^\frac{2}{3})}\)

[LaTex edited in by peter...not sure if I preserved Tim's order's of operations correctly (lots of parentheses), so see his top section for official version] – Tim Ye · 4 years, 7 months ago

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I don't mean to be intrusive, but I edited in a LaTex attempt of your answer so that people can read it easier. If that is not what you meant, feel free to re-edit it, using my LaTex and the formatting guide for help.

There is a link to the formatting guide at the base of the window where you enter a comment. Please put all math in LaTex as it is way easier to read than teasing out terms bounded by parentheses within parentheses. LaTex is actually easier, or just as hard as typing math the ugly way, but it is still easy to bungle. Please use the edit post button and the preview function, until your math states what you mean. Cheers:) – Peter Taylor Staff · 4 years, 7 months ago

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