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Mathematics 14 Online
OpenStudy (anonymous):

Solve for x by completing the squares: ax^2-x+2=0 Please show all the steps.

OpenStudy (anonymous):

really? this is icky

OpenStudy (anonymous):

\[ax^2-x=-2\] \[x^2-\frac{1}{a}x=-\frac{2}{a}\] \[(x-\frac{1}{2a})^2=-\frac{2}{a}+\frac{1}{4a^2}\] \[(x-\frac{1}{2a})^2=\frac{1-8a}{4a^2}\]

OpenStudy (anonymous):

take the square root of both sides, get \[x-\frac{1}{2a}=\pm\frac{\sqrt{1-8a}}{2a}\] then solve for \(x\) \[x=\frac{1\pm\sqrt{1-8a}}{2a}\]

OpenStudy (anonymous):

@Mertsj please don't tell me i screwed this up

OpenStudy (anonymous):

Thank you so much!

OpenStudy (mertsj):

yw

OpenStudy (anonymous):

no i think i added the fractions correctly

OpenStudy (anonymous):

\[\frac{1}{4a^2}-\frac{2}{a}=\frac{1-8a}{4a^2}\]

OpenStudy (mertsj):

yep. Somewhere I dropped the denominator fron the 2. Just getting tooooo old.

OpenStudy (anonymous):

wait until you get my age...

OpenStudy (mertsj):

I'm sure I'm older than you already.

OpenStudy (anonymous):

i would not bet on it

OpenStudy (mertsj):

Does that 73 mean something?

OpenStudy (anonymous):

but we will leave it a mystery

OpenStudy (anonymous):

well yes, but not my age !!

OpenStudy (anonymous):

it is satellite73 as in what i drive

OpenStudy (mertsj):

Very cool.

OpenStudy (anonymous):

thanks i was trying to find a picture but they seem to be either hot rods or beaters, and mine is neither

OpenStudy (mertsj):

Did you restore it yourself?

OpenStudy (anonymous):

What if the equation is like this: x^2+ax-2=0?

OpenStudy (mertsj):

It is easier because the coefficient of the x^2 term is already 1.

OpenStudy (mertsj):

Just take 1/2 of a, square it and add it to both sides. First, of course move the 2 to the right side.

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