Solve for x by completing the squares: ax^2-x+2=0 Please show all the steps.
really? this is icky
\[ax^2-x=-2\] \[x^2-\frac{1}{a}x=-\frac{2}{a}\] \[(x-\frac{1}{2a})^2=-\frac{2}{a}+\frac{1}{4a^2}\] \[(x-\frac{1}{2a})^2=\frac{1-8a}{4a^2}\]
take the square root of both sides, get \[x-\frac{1}{2a}=\pm\frac{\sqrt{1-8a}}{2a}\] then solve for \(x\) \[x=\frac{1\pm\sqrt{1-8a}}{2a}\]
@Mertsj please don't tell me i screwed this up
Thank you so much!
yw
no i think i added the fractions correctly
\[\frac{1}{4a^2}-\frac{2}{a}=\frac{1-8a}{4a^2}\]
yep. Somewhere I dropped the denominator fron the 2. Just getting tooooo old.
wait until you get my age...
I'm sure I'm older than you already.
i would not bet on it
Does that 73 mean something?
but we will leave it a mystery
well yes, but not my age !!
it is satellite73 as in what i drive
Very cool.
thanks i was trying to find a picture but they seem to be either hot rods or beaters, and mine is neither
Did you restore it yourself?
What if the equation is like this: x^2+ax-2=0?
It is easier because the coefficient of the x^2 term is already 1.
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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