Question

x^@-2x=-2 answwers will be in comments

Answer 1

a) x=-1+-i sqrt{2} b) x=1+- i sqrt{2} c) x=-1+-i d)x=1+-i

Answer 2

@hartnn

Answer 3

its, \(x^2-2x=-2\)
right ?

Answer 4

yes that person that was trying to help me on the other question confused me you helped lol but yes that is the equation

Answer 5

write it as \(x^2-2x+2=0\)
Compare ^ quadratic equation with \(ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two roots of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 6

a=1 b=-2 c=2

Answer 7

correct ,
b^2-4ac =... ?

Answer 8

-2^2-4(1)(2)=-4

Answer 9

good! :)
so, now just plug in the values in the formula

Answer 10

..... watttttt

Answer 11

\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 12

oh

Answer 13

can you solve firther ?

Answer 14

\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}=\frac{2 \pm \sqrt{-4}}{2}=\frac{2 \pm 2\sqrt{-1}}{2}\)
you know \(\sqrt{-1}=i\)
?

Answer 15

no not really this is where i get confused
and yes i know the sqrt of-1 is i

Answer 16

\(\huge \frac{2 \pm 2\sqrt{-1}}{2}=\frac{2 \pm 2i}{2}=\dfrac{2}{2}\pm\dfrac{2i}{2}\)
got this ?

Answer 17

yes so it should be d right

Answer 18

yes :D

Answer 19

yay im still gonna need your help on a few more questions

Answer 20

sorry, i have to leave, i need to go somewhere urgently....

Answer 21

oh ok