anonymous
  • anonymous
who knows how to COMPLETE THE SQUARE?!
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • 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.
anonymous
  • 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?
anonymous
  • anonymous
Maybe in about an hour. I'm running late. It depends on how this site goes with my iPhone too.

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anonymous
  • 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,\[c=b^2=\frac{81}{4}\]
anonymous
  • anonymous
So your equation is \[t^2-9t+\frac{81}{4}=({\pm}t{\mp}\frac{9}{2})^2\]
anonymous
  • anonymous
Let me know how you go.
radar
  • radar
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.
anonymous
  • anonymous
radar? do u think u can help hmoazeb?
anonymous
  • anonymous
ok thanks i will look at it,
anonymous
  • anonymous
thanks LOKISAN!
anonymous
  • anonymous
No worries :)

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