Question

What is the third step in solving this equation by completing the square?
9X^2 - 6x + 8 = 0

Answer 1

third step?

Answer 2

like :
\[9(x^{2} - 2/3*x + 8/9) = 0\]
and after that :
\[9*( (x-1/3)^{2} + 8/9 - 1/9)\]
?

Answer 3

thanks!

Answer 4

you're welcome

Answer 5

do not know what it means by third step

Answer 6

i do not*

Answer 7

you can do several things for your third step

Answer 8

oh, well thanks for all your help!

Answer 9

you forgot a square i think when taking the sqrt

Answer 10

yep
but i corrected it in the next step lol

Answer 11

yep :)

Answer 12

i noticed it to after i posted
i never check myself
zarkon knows

Answer 13

your 8 should be a 7

Answer 14

ah yes i ve a 7 too :/

Answer 15

omg you know what i added something to one side and i forgot to add it to the other

Answer 16

\[9(x^2-\frac{6}{9}x)+8=0\]
\[9(x^2-\frac{2}{3}x)+8=0\]
\[9(x^2-\frac{2}{3}x+(\frac{2}{2\cdot 3})^2)+8=9\cdot (\frac{2}{2 \cdot 3})^2\]
\[9(x^2-\frac{2}{3}x+(\frac{1}{3})^2)+8=9\cdot (\frac{1}{3})^2\]
\[9(x-\frac{1}{3})^2+8=9 \cdot \frac{1}{9}\]
\[9(x-\frac{1}{3})^2+8=1\]
\[9(x-\frac{1}{3})^2=1-8\]
\[9(x-\frac{1}{3})^2=-7\]
\[3(x-\frac{1}{3})=\pm \sqrt{-7}\]
\[3(x-\frac{1}{3})=\pm i \sqrt{7}\]
\[x-\frac{1}{3}=\pm \frac{i \sqrt{7}}{3}\]
\[x=\frac{1}{3} \pm \frac{i \sqrt{7}}{3}\]

Answer 17

there!

Answer 18

lol

Answer 19

i'm gonna eat