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I'm trying to solve a more complex cubic equation but to simplify things as a start I picked this one:

$$ 3\cdot 4^3+2\cdot 4-200=0 $$

Here $x$ is $4$.

I'm looking at wikipedia and trying to solve with the general cubic formula.

In my case I have

$$ a = 3, b = 1, c = 2, d = -200 $$

So solving for $\Delta_0$

$$ \Delta_0 = b^2 - 3ac $$ $$ \Delta_0 = 1^2 - 3\cdot3\cdot2 $$ $$ \Delta_0 = -17 $$

Solving for $\Delta_1$

$$ \Delta_1 = 2b^3 - 9abc + 27a^2d $$ $$ \Delta_1 = 2\cdot1^3 - 9\cdot3\cdot1\cdot2 + 27\cdot3^2\cdot-200 $$ $$ \Delta_1 = −48652 $$

At this point already I feel like something is wrong.

I kept going and got some very negative numbers for the rest, which are obviously not correct, so I'll spare you the details.

What am I doing wrong?

FishDrowned
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php_nub_qq
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    $b=0$, not $1$. There is no quadratic term in $x$. – lulu Dec 05 '23 at 19:30
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    The equation is $3x^3+2x-200=(3x^2 + 12x + 50)(x - 4)=0$. The the cubic is reducible, i.e., a linear equation and a quadratic one. No need to use a formula for a cubic. – Dietrich Burde Dec 05 '23 at 19:30
  • @lulu but this doesn't seem to make much difference.. I'll re-calculate with b=0 and will update the question. – php_nub_qq Dec 05 '23 at 19:32
  • It makes a difference, because now the equation is just quadratic, after dividing by $x-4$. Why would you use the cubic formula for it. – Dietrich Burde Dec 05 '23 at 19:33
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    Just to say, though it isn't necessary to use Cardano's methods for this cubic, as it is reducible, you certainly can. For comparison, using Wikipedia's notation, I get $\Delta_0=-18$, $\Delta_1=-48600$, $C=0.493242009$ (I took the $+$ square root), which do indeed give $x=4$. – lulu Dec 05 '23 at 19:50
  • @lulu how did you get that for C? I'm getting $22.989245129−22.989282972i$ .. – php_nub_qq Dec 05 '23 at 20:22
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    $C$ is clearly real. The only thing that could make it complex is the term $\sqrt {\Delta_1^2-4\Delta_0^3}$. But $\Delta_1$ is huge compared to $\Delta_0$, so that term obviously dominates, making the square root real. – lulu Dec 05 '23 at 20:24
  • Correction: Since $\Delta_0<0$ both of the terms inside the square root are positive, making it even more clear that $C$ is real. – lulu Dec 05 '23 at 20:33
  • @lulu thank you! It appears that the error was all in how I was inputting negative numbers in my calculator, apparently it requires them to be in braces otherwise it multiplies first, then assigns the negative sign, i.e. $-2^2$ would give $-4$, needs to be $(-2)^2$. Well that was silly, anyway with it out of the way and the correction of $b=0$ I got the expected result. – php_nub_qq Dec 05 '23 at 20:38
  • Excellent. $\quad$ – lulu Dec 05 '23 at 20:41

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