1. steve816

Solve in the complex number system: $x^2+3=x$

2. zzr0ck3r

$$x^2-x+3=0\implies (x-\dfrac{1}{2})^2=-3+(\dfrac{1}{2})^2\\(x-\dfrac{1}{2})^2=\dfrac{-11}{4}$$

3. zzr0ck3r

Can you complete here?

4. steve816

Is completing the square the only method or was I allowed to use the quadratic formula?

5. steve816

I can finish off from here, thanks for the help :)

6. zzr0ck3r

sure, you can use that. This was just faster...

7. zzr0ck3r

The quadratic equation is derived using the "completing the square" method :)

8. steve816

Oh really? Never really understood how the quadratic equation worked even though i'm in pre-calculus.

9. zzr0ck3r

$ax^2+bx+c=0\\a(x^2+\frac{b}{a}x)=-c\\a(x+\dfrac{b}{2a})^2=-c+a\dfrac{b^2}{4a^2}\\(x+\dfrac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\(x+\dfrac{b}{2a})^2=\dfrac{b^2-4ac}{4a^2}\\(x+\dfrac{b}{2a})=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\\(x+\dfrac{b}{2a})=\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

10. steve816

Holy guacamole! :o *jaw drops*

11. wolf1728

12. zzr0ck3r

pretty much the same way.

13. wolf1728

steve - did you check that answer?

14. wolf1728

15. steve816

Yup, I got the answer right.

16. anonymous

17. steve816

18. wolf1728

I don't get that answer a=1 b=-1 c=3 right?

19. anonymous

Yeah

20. wolf1728

x = 1 +- sq root(1 -4*1*3) / 2 x = [1 +- sq root(-11)] / 2

21. steve816

yeah, that is what I got.

22. wolf1728

Okay, it's just that I didn't see that answer printed anywhere.

23. steve816