Question

Someone please solve this for me.
solve by completing the square.
3x^2-6x=17
idk wat to do after (3x+3)^2=27 (not even sure if that is correct)

Answer 1

you cannot complete the square that way.

Answer 2

How come? If that is the case, would it be better to factor this problem??

Answer 3

the leading coefficient has to be one, so you have to divide by 3 first

Answer 4

first take out 3
3(x^2 - 2x) = 17

Answer 5

\[3x^2-6x=17\]
\[x^2-2x=\frac{17}{3}\]
\[(x-1)^2=\frac{17}{3}+1=\frac{20}{3}\]
\[x-1=\pm\sqrt{\frac{20}{3}}\]
\[x=1\pm\sqrt{\frac{20}{3}}\]

Answer 6

(x^2 - 2x) = 17/3

satellite has it

Answer 7

you can factor out a 3 if you like, but you have an equation so you are allowed to divide both sides by 3 to make the leading coefficient one