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completing the square

wellyn

New member
Dec 15, 2013
2
help im stumped on this ((x2)/18)-(x/9)=1(Headbang)(Headbang)(Headbang)(Headbang)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: competing the square

help im stumped on this ((x2)/18)-(x/9)=1(Headbang)(Headbang)(Headbang)(Headbang)
We are given:

\(\displaystyle \frac{x^2}{18}-\frac{x}{9}=1\)

I think I would first multiply through by the lowest common denominator to get rid of the denominators. So, multiplying through by 18, we get:

\(\displaystyle x^2-2x=18\)

Can you proceed?
 

wellyn

New member
Dec 15, 2013
2
Re: competing the square

no sorry im stumped
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: competing the square

no sorry im stumped
You want to take half the coefficient of the linear term (the term with $x$ as a factor) and square it, and add this to both sides. What do you get?