## anonymous 5 years ago who knows how to COMPLETE THE SQUARE?!

1. anonymous

Me! If you have the quadratic,$ax^2+bx+c$factor out the a, and do the following:$a(x^2+\frac{b}{a}x+\frac{c}{a})=a(x^2+\frac{b}{a}x+(\frac{b/a}{2})^2-(\frac{b/a}{2})^2+\frac{c}{a})$$=a[(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2]$$=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$If your quadratic is set to zero (i.e. you're finding the roots), then$a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}=0\rightarrow (x+\frac{b}{2a})^2=\frac{\frac{b^2}{4a}-c}{a}$Ssquare toot both sides and subtract b/2a to yield x.

2. anonymous

ok so here is my problem... t^2 - 9t _____?____ i need to first make a completed trinomial and then that equals a binomial squared. can you help me through this?

3. anonymous

Maybe in about an hour. I'm running late. It depends on how this site goes with my iPhone too.

4. anonymous

I have your answer - just came out of the shower and did it (I shortened my shower for you!). You have the following setup:$t^2-9t+c=(at+b)^2$Don't get these a, b, c's mixed up with what I wrote above - I just picked some pronumerals. Expand the RHS and compare with the LHS - two polynomials are equal if and only if the coefficients of the respective powers are equal, so$t^2-9t+c=a^2t^2+2abt+b^2$implies$a^2=1 \rightarrow a=\pm1$which means for the 't' term,$2(\pm1)b=-9 \rightarrow b={\mp}\frac{9}{2}$(pay attention to the signs here). Therefore,$c=b^2=\frac{81}{4}$

5. anonymous

So your equation is $t^2-9t+\frac{81}{4}=({\pm}t{\mp}\frac{9}{2})^2$

6. anonymous

Let me know how you go.

ok so here is my problem... t^2 - 9t _____?____ so you are to choose a number to make the results well it looks like lokisan pretty much took care of it.

8. anonymous

radar? do u think u can help hmoazeb?

9. anonymous

ok thanks i will look at it,

10. anonymous

thanks LOKISAN!

11. anonymous

No worries :)