Question

Simple question about equation:

Answer 1

\[z3=\frac{ -2-2j+j ^{2} }{ 2+2j-j ^{2} }\]

Answer 2

What I must do here?

Answer 3

is j the imaginary number , j^2=-1 ?

Answer 4

yes

Answer 5

go ahead and simplify , replace j^2 with -1

Answer 6

Ok but I have the same difficult with x when I havent j and I HAVE X

Answer 7

Ok I change it

Answer 8

\[z3=\frac{ -2-2x+x ^{2} }{ 2+2x-x ^{2} }\]

Answer 9

with this thing I have problems

Answer 10

z_3 = (-2 -2j + j^2) / ( 2 + 2j -j^2)
= ( -2 -2j -1 ) / ( 2 + 2j - (-1) )
= ( -3 -2j) / ( 3 +2j)
= -(3 + 2j) / (3 + 2j)
= -1

Answer 11

you multiplicate with -1?

Answer 12

with x you can try factoring it

Answer 13

I factored out -1 from (-3 -2j )
(-3 -2j ) = - ( 3 + 2j)

Answer 14

Ok I cant understan how function this factor

Answer 15

The numerator is given as (-2 -2x +x^2)
but
-1( 2 +2x -x^2) = -2 -2x + x^2
check it for yourself, distribute

Answer 16

a ok I understand and with x?

Answer 17

I dont understand because I cant simplyfi with x

Answer 18

Hi

Answer 19

what do you mean simplify? you can factor out -1 , and then cancel

Answer 20

you mean factor?

Answer 21

it is not necessary to factor the quadratic

Answer 22

\[z3=\frac{ -2-2x+x ^{2} }{ 2+2x-x ^{2} }\]

Answer 23

replace the numerator -2 -2x + x^2 with -( 2 + 2x -x^2)

Answer 24

when I have this I cant simplyfi

Answer 25

I cant understand why I cant simplify +x^2 with x2 and other number

Answer 26

x^2/-x^2

Answer 27

do you agree that
-2 -2x + x^2 = - ( 2 + 2x -x^2)

Answer 28

because of order of operations. you cannot divide before adding all the terms in the numerator

Answer 29

ok and wher I can find other exercis of this? what I must search on google beacuse I have difficulty

Answer 30

\[z3=\frac{ -2-2x+x ^{2} }{ 2+2x-x ^{2} } ~~ I ~~ mean ~~ other ~~ exercise~~ like ~~ this\]

Answer 31

can you finish the problem simplifying