Question

x^2+√3x-1+ how do I solve this?

Answer 1

\[x^2+\sqrt{3}x-1=0\] like that?

Answer 2

yes

Answer 3

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=1,b=\sqrt{3},c=-1\]

Answer 4

get
\[x=\frac{-\sqrt{3}\pm\sqrt{3-4\times 1\times -1}}{2}\]

Answer 5

or
\[x=\frac{-\sqrt{3}\pm\sqrt{7}}{2}\]

Answer 6

so the square root under the square root cancels out?

Answer 7

? nothing cancels

Answer 8

maybe you mean when you compute
\[b^2\] with
\[b=\sqrt{3}\] you get
\[\sqrt{3}^2=3\] which is right

Answer 9

yeah duh, I freaked out for no reason. Thanks!!!

Answer 10

yw