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

third step?

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

thanks!

you're welcome

do not know what it means by third step

i do not*

you can do several things for your third step

oh, well thanks for all your help!

you forgot a square i think when taking the sqrt

yep but i corrected it in the next step lol

yep :)

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

your 8 should be a 7

ah yes i ve a 7 too :/

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

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

there!

lol

i'm gonna eat

Join our real-time social learning platform and learn together with your friends!