Solve:

x^3 + 6x = 4x^2

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- Apr 10th 2009, 10:51 AMMr RayonSolve a cubic equation
Solve:

x^3 + 6x = 4x^2 - Apr 10th 2009, 10:57 AMstapel
Follow the standard process: Move everything over to one side of the "equals" sign, take

**the common factor**out front, and then apply**the Quadratic Formula**to solve the remaining quadratic factor.

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink) - Apr 10th 2009, 11:03 AMMr Rayon
- Apr 10th 2009, 11:11 AMMr Rayon
Yes...I get what your saying now. Ok...I'll try that.

- Apr 10th 2009, 11:32 AMTKHunny
- Apr 10th 2009, 07:10 PMMr Rayon
x^3 + 6x = 4x^2

x^3 - 4x^2 + 6x = 0

x(x^2 - 4x + 6) = 0

x= (-(-4)±√((-4)^2 -4(1)(6)))/2

x= (4±√(16 -24))/2

x= (4±√-8)/2

What did I do wrong? I can't take the square root of a negative number. And we haven't gone through imaginary numbers yet so I don't think we would you those to solve the above. - Apr 10th 2009, 07:16 PMe^(i*pi)
- Apr 10th 2009, 07:16 PMmr fantastic
- Apr 11th 2009, 05:51 AMstapel
If your class hasn't covered complex numbers yet, then you haven't done anything "wrong"; you've accurately demonstrated that there is only one

*real-number*solution. (Wink)

(They were supposed to have mentioned this potential issue in your class and your textbook!)