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Mathematics 18 Online
OpenStudy (anonymous):

solve by completing the square m^2-49/2m=1/2

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..

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):

\[\huge (x-3) (x+7) \]

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

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

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

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

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?

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