Question

if 3x^2-2x+7=0 then (x-1/3)^2=
A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9

Answer 1

hmm I don't see the relation
what section is this covering?

Answer 2

almost sounds like saying

if potatoes are round
then is going to rain on thailand

which are two unrelated elements

Answer 3

true I see it that way too xD I dont know the section thats why I post it here

Answer 4

go eat a chapstick

Answer 5

@PleaseHelp0204 who are you saying thast and why? xd

Answer 6

well i just ate 4 and now i dont know what im doing

Answer 7

i feel dizzy

Answer 8

hahahahaha xDDD I eat weird stuff when im bored but not to that point xD

Answer 9

Lol i did enjoy that @jdoe0001

So basically what i see, is a polynomial *that has roots i would assume we need to solve for*

And what we get from that result....we plug those results into the 2nd expression given for 'x' and solve that

Answer 10

If 3x2 - 2x + 7 = 0 then x2-(2/3)x+7/3=0
In solving by completion of the square we have
(x-1/3)2=x2-(2/3)x+1/9 The equation is
x2-(2/3)x++21/9=0 and so
(x-1/3)2+20/9=0 and so (x-1/3)2=-20/9. No need to calculate the roots at all.

Answer 11

If its not asking for anything? xd

Answer 12

True, i like the roots though XD

But yes correct answer

Answer 13

I dont get anything xd

Answer 14

now that i am looking at what i did. i dont get it either lol... must be the chapsticks

Answer 15

hahahahahaha xDDDD

Answer 16

help john

Answer 17

i have no idea

Answer 18

Lol oh no no you work through the pain of those chapsticks and you explain!! XD

Answer 19

all right well kfsv do you know completion of the squares?

Answer 20

i think you can also use the quadratic formula

Answer 21

Quadratic formula would probably be easiest.

(2 + / - sqrt (4 - 84)) / 6
(2 + / - sqrt (-80)) / 6
(2 + / - 4i(sqrt5) / 6
x=(1 + 2i(sqrt5)) / 3 OR (1 - 2i(sqrt5)) / 3

((1 + 2i(sqrt5)) / 3 - 1/3)^2 = (((2i(sqrt5))/3)^2) = -20/9
((1 - 2i(sqrt5))/3 - 1/3)^2 = (((-2i(sqrt5))/3)^2) = -20/9

-20/9

Answer 22

john u think thats better

Answer 23

thats too complicated D: