Mathematics
OpenStudy (anonymous):

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

myininaya (myininaya):

third step?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

you're welcome

myininaya (myininaya):

do not know what it means by third step

myininaya (myininaya):

i do not*

myininaya (myininaya):

you can do several things for your third step

OpenStudy (anonymous):

oh, well thanks for all your help!

OpenStudy (anonymous):

you forgot a square i think when taking the sqrt

myininaya (myininaya):

yep but i corrected it in the next step lol

OpenStudy (anonymous):

yep :)

myininaya (myininaya):

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

OpenStudy (zarkon):

your 8 should be a 7

OpenStudy (anonymous):

ah yes i ve a 7 too :/

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

there!

myininaya (myininaya):

lol

myininaya (myininaya):

i'm gonna eat

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