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

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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

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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

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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

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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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