Question

Factor x^2-x-1?

Answer 1

go get 'em myininaya!

Answer 2

still working?

Answer 3

\[b=(\frac{-1}{2}+z)+(\frac{-1}{2}-z)\]
\[a*c=-1=(\frac{-1}{2}+z)(\frac{-1}{2}-z)=\frac{1}{4}-z^2=>z^2=\frac{1}{4}-ac=>z=\pm \sqrt{\frac{1}{4}-ac}\]
replace bx with \[bx=(\frac{-1}{2}+\sqrt{\frac{1}{4}-ac})x+(\frac{-1}{2}-\sqrt{\frac{1}{4}-ac})x\]
so we have
\[x^2+(\frac{-1}{2}+\sqrt{\frac{1}{4}-ac})x+(\frac{-1}{2}-\sqrt{\frac{1}{4}-ac})x-1\]
but ac=-1
\[x^2+(\frac{-1}{2}+\sqrt{\frac{1}{4}-(-1)})x+(\frac{-1}{2}-\sqrt{\frac{1}{4}-(-1)})x-1\]
\[x^2+(\frac{-1}{2}+\sqrt{\frac{1+4}{4}})x+(\frac{-1}{2}-\sqrt{\frac{1+4}{4}})x-1\]
\[x^2+(\frac{-1+\sqrt{5}}{2})x+(\frac{-1-\sqrt{5}}{2})x-1\]
\[x(x+\frac{-1+\sqrt{5}}{2})+\frac{-1-\sqrt{5}}{2}(x-\frac{2}{-1-\sqrt{5}})\]
\[x(x+\frac{-1+\sqrt{5}}{2})+\frac{-1-\sqrt{5}}{2}(x+\frac{2}{1+\sqrt{5}}*\frac{1-\sqrt{5}}{1-\sqrt{5}})\]
\[x(x+\frac{-1+\sqrt{5}}{2})+\frac{-1-\sqrt{5}}{2}(x+\frac{2(1-\sqrt{5})}{1-5})\]
\[x(x+\frac{-1+\sqrt{5}}{2})+\frac{-1-\sqrt{5}}{2}(x-\frac{1-\sqrt{5}}{2})\]
\[x(x+\frac{-1+\sqrt{5}}{2})+\frac{-1-\sqrt{5}}{2}(x+\frac{-1+\sqrt{5}}{2})\]
\[(x+\frac{-1+\sqrt{5}}{2})(x+\frac{-1-\sqrt{5}}{2})\]

Answer 4

Wow. I believe you since when this equation = 0, x = \[\left( 1\pm \sqrt{5} \right) \div2\]. Impressive!

Answer 5

lol

Answer 6

are you out of your mind?

Answer 7

he said impressive

Answer 8

impressed with how long to took to get
\[x=\frac{1\pm\sqrt{5}}