# Solve a cubic equation

• Apr 10th 2009, 10:51 AM
Mr Rayon
Solve a cubic equation
Solve:

x^3 + 6x = 4x^2
• Apr 10th 2009, 10:57 AM
stapel
Quote:

Originally Posted by Mr Rayon
Solve: x^3 + 6x = 4x^2

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 AM
Mr Rayon
Quote:

Originally Posted by stapel
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)

But it's a cubic. I don't know how to use the Quadratic Formula for cubic expressions. I've only used it for quadratic expressions.
• Apr 10th 2009, 11:11 AM
Mr Rayon
Yes...I get what your saying now. Ok...I'll try that.
• Apr 10th 2009, 11:32 AM
TKHunny
Quote:

Originally Posted by Mr Rayon
I don't know how to use the Quadratic Formula for cubic expressions. I've only used it for quadratic expressions.

This is a wonderful statement. Please remember this lesson and never even consider using the Quadratic Formula on a Cubic. Make sure it's quadratic in some way shape or form before trying.
• Apr 10th 2009, 07:10 PM
Mr 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 PM
e^(i*pi)
Quote:

Originally Posted by Mr 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.

It means your only real solution is x=0 from the factor of x you too out at the start
• Apr 10th 2009, 07:16 PM
mr fantastic
Quote:

Originally Posted by Mr 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.

x = 0 is the only real solution to the equation you posted.
• Apr 11th 2009, 05:51 AM
stapel
Quote:

Originally Posted by Mr Rayon
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.

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!)