what are the real or imaginary solutions of the polynomial equation 4x3+4=0

Question

mertsj · Answer

1.  Subtract 4 from both sides
2.  Divide both sides by 4
3.  Take the 3rd root of  both sides.

anonymous · Answer

whats the answer @Mertsj

mertsj · Answer

http://www.wolframalpha.com/input/?i=Solve%3A++4x^3%2B4%3D0

anonymous · Answer

thankkk Uu @Mertsj

zepdrix · Answer

If you need exact solutions, you can apply your sum of cubes formula, $$\Large\bf\sf 4(x^3+1)=0,\qquad\qquad x^3+1=0$$$$\Large\bf\sf (x+1)(x^2-x+1)=0$$then follow it up with the quadratic formula.

zepdrix · Answer

no big woop, just another option :D

anonymous · Answer

thanks but i have to turn this in tomoorow and i have a bad grade if u could help me with the answers plz @zepdrix

zepdrix · Answer

$$\Large\bf\sf (x+1)(\color{royalblue}{x^2-x+1})=0$$Apply quadratic formula to the blue part.$$\Large\bf\sf \color{royalblue}{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$
Plug stuff in,
$$\Large\bf\sf \color{royalblue}{x=\frac{1\pm\sqrt{1-4}}{2}}$$Simplify.