katherinesmith
  • katherinesmith
solve for m. picture inside! please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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katherinesmith
  • katherinesmith
\[\frac{ 1 }{ m } = \frac{ m - 34 }{2m ^{2} }\]
DebbieG
  • DebbieG
You could cross-multiply to eliminate the fractions. OR multiply both sides by the LCD - either method will work. Then you won't have any fractions Then you collect all the terms on one side, set =0. It will be a (really simple) quadratic that you can solve by factoring.
katherinesmith
  • katherinesmith
okay that didn't help me at all because I don't know how to cross multiply those kinds of numbers or even find the LCD.

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More answers

DebbieG
  • DebbieG
Do you know how to multiply m*m?
katherinesmith
  • katherinesmith
m^2
DebbieG
  • DebbieG
Good. Do you know how to multiply m(m-34) ?
katherinesmith
  • katherinesmith
m^2 - 34m
DebbieG
  • DebbieG
Excellent. Do you know how to multiply 2m^2 * (1) ? :)
katherinesmith
  • katherinesmith
so now I have 2m^2 = m^2 - 34m. now what.
katherinesmith
  • katherinesmith
then I can get m^2 = -34m
DebbieG
  • DebbieG
Collect al the terms on one side. I'd suggest moving the terms to the left side, just to make sure that the leading coefficient is positive.
DebbieG
  • DebbieG
Yes, exactly! Now bring that -34m on over to the LHS.... and you get?
katherinesmith
  • katherinesmith
34m + m^2 = 0 ?
DebbieG
  • DebbieG
Sure,t hat will work (although it's "standard" to put the x^2 term first, then the x term... but it won't make any real difference here).
katherinesmith
  • katherinesmith
but isn't there another step
DebbieG
  • DebbieG
OK, so you have now: \(\Large 34m+m^2=0\) That's a lovely little quadratic equation. Do you have any ideas on how to solve that equation? What do you do next?
DebbieG
  • DebbieG
Oh yes, we aren't done yet. :)
phi
  • phi
factor out an "m" from each term
katherinesmith
  • katherinesmith
m(m + 33m) = 0?
DebbieG
  • DebbieG
Why did you change the 34 to 33? m * 33m = 33m^2, not 34m^2.
phi
  • phi
factor is the opposite of distribute. notice m(m+33m) will not give you the original
katherinesmith
  • katherinesmith
okay then what am I supposed to do.
DebbieG
  • DebbieG
Wait, did you fix your factoring step?
katherinesmith
  • katherinesmith
im stuck at m^2 + 34m = 0. don't know how to factor it out correctly.
phi
  • phi
write m^2 +34 m as m*m + 34*m notice both terms are multiplied by m, so you can factor one "m" out
DebbieG
  • DebbieG
Here is an example with different numbers: \[\Large 25x+x^2=x(25 + x)\]
katherinesmith
  • katherinesmith
okay so m(m + 34) = 0
phi
  • phi
if m=0 what is m(m + 34)= to ? if (m+34) were 0 what is m(m + 34)= to ?
DebbieG
  • DebbieG
Good. so there is a rule called the Zero Factor Property. It says that: If A*B=0 then A=0 or B=0.
katherinesmith
  • katherinesmith
s... m = 0 or 34 = 0 ?
katherinesmith
  • katherinesmith
m = 0 or m + 34 = 0 ?
DebbieG
  • DebbieG
Right!
DebbieG
  • DebbieG
so either m=0, or m = ??
katherinesmith
  • katherinesmith
34
katherinesmith
  • katherinesmith
or is it -34
DebbieG
  • DebbieG
You tell me? :) It is one and not the other.
katherinesmith
  • katherinesmith
I don't know!
DebbieG
  • DebbieG
m+34 = 0 What value of m makes that true?
katherinesmith
  • katherinesmith
-34
DebbieG
  • DebbieG
Well, does m=34 make it true?
katherinesmith
  • katherinesmith
no only -34
DebbieG
  • DebbieG
RIGHT! See, stop selling yourself short. You DO know! :) Just take a deep breath and think about it! :)
katherinesmith
  • katherinesmith
thank you for your help.
DebbieG
  • DebbieG
NOW, you have 2 solutions: m=0 or m=-34 But there is just one little wrinkle here.....
DebbieG
  • DebbieG
THOSE are solutions to the equation: \[\Large 34m+m^2=0\] Which we obtained by cross-multiplying the ORIGINAL equation. The original equation involves rational expressions. And with rational expressions, you always have to make sure that they are WELL DEFINED, which basically just means that you don't have a 0 in a denominator So ALWAYS check your solutions to make sure that none of them give you a zero denominator. If so, that is an EXTRANEOUS SOLUTION... meaning that it ISNT a solution to the original equation. So just throw it out, and keep any that aren't extraneous.

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