Mathematics
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OpenStudy (anonymous):
solve by completing the square m^2-49/2m=1/2
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OpenStudy (anonymous):
ok that last problem we did it says that there is not approximate answer because it is all integers
OpenStudy (saifoo.khan):
OIC.
OpenStudy (saifoo.khan):
well, this prob, cant get the ques! :(
OpenStudy (anonymous):
\[m^2 - (49/2)*m = 1/2\]
OpenStudy (anonymous):
well hello hero..
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OpenStudy (anonymous):
OK, you want to turn this into the form of \[(a+b)^2\] corret?
OpenStudy (anonymous):
yeah
OpenStudy (anonymous):
correct
OpenStudy (anonymous):
is this the original equation or have you started some of the steps?
OpenStudy (saifoo.khan):
\[\huge (x-3) (x+7) \]
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OpenStudy (saifoo.khan):
this?
OpenStudy (anonymous):
well the procedure is to take the coefficient on the x term, divide it by 2, then square that value and add it to both sides
OpenStudy (anonymous):
ok, I see what your saying
OpenStudy (anonymous):
so \[a*x^2 + b*x + c = d\]
then \[(b/2)^2\] would be added to both sides as such
\[a*x^2 + b*x + (b/2)^2 = d-c+(b/2)^2\]
OpenStudy (anonymous):
that was the equation
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hero (hero):
the answer is \[\frac {\pm \sqrt{2409} + 49} {4} \]
OpenStudy (anonymous):
whoops! added a term
OpenStudy (saifoo.khan):
\[\huge \color\purple (x+2)^2-25 \]
OpenStudy (anonymous):
so you have \[m^2 + (49/2)*m = 1/2\] would be \[((49/2)/2)^2\] as the term to add to both sides
hero (hero):
oops m = 0, 49
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OpenStudy (anonymous):
\[m^2 + (49/2) * m + (49/4)^2\] would be \[(m+49/4)^2\]
OpenStudy (anonymous):
that was the equation
OpenStudy (anonymous):
thats confusing
OpenStudy (anonymous):
\[(m+49/4)^2 = (1/2) + (49/4)^2\]
OpenStudy (anonymous):
and then you can solve for m
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OpenStudy (saifoo.khan):
\[\Huge WAYYYYY-CONFUSING!\]
hero (hero):
It's not confusing
hero (hero):
I actually like completing the square better than using quadratic formula
OpenStudy (anonymous):
that was the equation
OpenStudy (anonymous):
? not if you are good with numbers.. I agree than
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hero (hero):
It's a more algebraic way of solving for m
OpenStudy (anonymous):
yeah, it seems more like a way to familiarize you with factoring and understanding quadratic functions
OpenStudy (anonymous):
was this clear? at which point did you lose track?