Question

Please help, I'm stuck at 4 AM doing math!

Answer 1

Solve in the complex number system:
\[x^2+3=x\]

Answer 2

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

Answer 3

Can you complete here?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Holy guacamole! :o *jaw drops*

Answer 11

Here's another explanation of deriving the quadratic formula:
http://www.1728.org/quadr2.htm

Answer 12

pretty much the same way.

Answer 13

steve - did you check that answer?

Answer 14

the (-11/4) answer?

Answer 15

Yup, I got the answer right.

Answer 16

The answer is (-11/4)

Answer 17

that wasn't the answer... @lauren.ascough

Answer 18

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

Answer 19

Yeah

Answer 20

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

Answer 21

yeah, that is what I got.

Answer 22

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

Answer 23

Thanks for the answer though.

Answer 24

Im stuck at 6AM playing and modding skyrim

Why?

BECAUSE IM IN VIRTUAL SCHOOL