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I have searched for the cubic formula, which is: $$ \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{\left(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}\right) ^ 2 + \left(\frac{C}{3A} - \frac{B^2}{9A^2}\right)^3}} + \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{\left(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}\right) ^ 2 -\left(\frac{C}{3A} - \frac{B^2}{9A^2}\right)^3}} - \frac{B}{3A}$$ However, when I substitute A = 1, B = 6, C = 11, D = 6, then I should expect something like this: $$ x^3 + 6x^2 + 11x + 6 = 0 $$ $$ x = -1 \, or \, x = -2 \, or \, x = -3 $$ However I have got complex solution. What is happening?

mathlove
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  • I do not understand what you are trying to ask... please be more precise.. –  Jan 06 '14 at 09:38
  • try to substitute the numbers as I have told and tell if there are any errors. –  Jan 06 '14 at 09:38
  • Please mind That is not my job to do it for you... That is your job to do and if you have some problem then some one could help it.. –  Jan 06 '14 at 09:39
  • I have done it and I have calculated the value but I got a complex solution. I should expect a real solution since there is no complex solution in the cubic equation $x^3+6x^2+11x+6=0$ –  Jan 06 '14 at 09:41
  • No No... It is not necessarily true that you will get all three roots as real... There is a possibility that one root is real and rest two are complex so you may not have done any mistake... –  Jan 06 '14 at 09:42
  • However for the example I have told, there should be no complex solution. But when I use the formula to solve the equation I got a complex solution. –  Jan 06 '14 at 09:43
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    No worry. If you substitute the number into the formula, even when all the roots are real, it is still very common you get an expression that involve complex numbers. However, if you evaluate the expression exactly, all the imaginary pieces will cancel out. – achille hui Jan 06 '14 at 09:44
  • You made a mistake while coping your formula. In your second cube root, there should be a minus sign before the square root and a plus sign inside the square root where you now have a minus sign. Therefore, you'll see that the cube roots cancel and you'll be left with the last term which is -2. – Raskolnikov Jan 06 '14 at 09:45
  • Do YOU get a complex solution, or does a computer get it for you ? I ask this because, if you are using a mathematical software, it is not uncommon for it to choose as value of the cube root any of other two complex cubic roots of the number in question. E.g., $\sqrt[3]1$ could be $1$, but it could also be $\dfrac{-1\pm i\sqrt3}2$ , for instance. – Lucian Jan 06 '14 at 23:23

1 Answers1

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You made a mistake while copying your formula. In your second cube root, there should be a minus sign before the square root and a plus sign inside the square root where you now have a minus sign. Therefore, you'll see that the cube roots cancel and you'll be left with the last term which is -2.

The correct formula:

$$\sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}) ^ 2 + (\frac{C}{3A} - \frac{B^2}{9A^2})^3}} + \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} - \sqrt{(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}) ^ 2 + (\frac{C}{3A} - \frac{B^2}{9A^2})^3}} - \frac{B}{3A}$$

Raskolnikov
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