Question

x^2-2/3x (same instructions) @satellite73

Answer 1

the fractions kind of scaring me

Answer 2

half of \(\frac{2}{3}\) is ?

Answer 3

rude

Answer 4

don't let them scare you
just take half of two thirds

Answer 5

hm 1/3?

Answer 6

not 1/3?
but rather 1/3!!

Answer 7

that is step one
step two is what is \(\left(\frac{1}{3}\right)^2\)

Answer 8

okay so 1/9

Answer 9

exactly

Answer 10

so
\[x^2-\frac{2}{3}x+\frac{1}{9}=(x-\frac{1}{3})^2\]

Answer 11

these seem a lot simpler than I thought. Thanks for helping me realize how to find the answer and how to not overthink things and doubt

Answer 12

they are very simple once you know the rules
really easy
even with (ugh) fractions

Answer 13

could we do one more?

Answer 14

ok why not?

Answer 15

so x^2-1/3x

Answer 16

ooh another fraction
want to try it?

Answer 17

half of 1/is 1/6 right? 1/3 x 1/2? (so much for not doubting)

Answer 18

right, half of \(\frac{1}{3}\) is \(\frac{1}{6}\)

Answer 19

then second , what is \((\frac{1}{6})^2\) ?

Answer 20

x^2-1/3+1/12= (x^2-1/6)2

Answer 21

oops
so close

Answer 22

you balked on \[\huge \left(\frac{1}{6}\right)^2\]

Answer 23

x^2 -1/3x+1/36=(x^2-1/6)^2 WOW..

Answer 24

yup

Answer 25

Thanks. Shall we complete the square now? I just need some help getting started

Answer 26

lol what the heck you think you have been doing???

Answer 27

that's what the second set of instructions is asking for specifically so I am going by what my textbook says

Answer 28

ok lets try one
believe me it is the same

Answer 29

okay let's do this
x^2+6x=7

Answer 30

ready?

Answer 31

you would still take hald of six and square it ?

Answer 32

yes

Answer 33

*half

Answer 34

subtract 7 over?

Answer 35

no

Answer 36

leave it right where it is

Answer 37

you want me to walk you though this one?

Answer 38

x^2+6x+9=7?

Answer 39

ok you started with
\[x^2+6x=7\] when you complete the square as you have been doing you get
\[x^2+6x+9\] but you have to add \(9\) to the right as well
so it is
\[x^2+6x+9=7+9\\
(x+3)^2=16\]

Answer 40

then it is a snap to solve
\[(x+3)^2=16\\x+3=4,x+3=-4\\
x=1,x=-7\]

Answer 41

i see said the blind man

Answer 42

to his deaf wife over the disconnected telephone as he stuck his head out of the closed window to see if it was raining

Answer 43

you can really go right from
\[x^2+6x=7\] to
\[(x+3)^2=7+9\] if you like

Answer 44

i am tired
want to try one more?

Answer 45

so for x^2-2x=2
x^2-2x+1=2+1
(x^2+2)^2=3?
I think I just confused myself

Answer 46

everything good but the last line

Answer 47

\[(x-1)^2=3\]

Answer 48

?

Answer 49

\[x^2-2x=2\\
x^2-2x+1=2+1\\
(x-1)^2=3\]

Answer 50

just like
\[x^2+6x=7\\
x^2+6x+9=7+9\\
(x+3)^2=16\]

Answer 51

just foiled out the (x-1)^2 part , I don't know why but it threw me off.
Anyway, I don't know how else to thank you, you've been very helpful. Also, sorry for hogging you @satellite73 goodnight!

Answer 52

hey no problem
gnight to you