anonymous
  • anonymous
18m^2-1 factor plz!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[81m^2-1\]
anonymous
  • anonymous
wait which one, or both
anonymous
  • anonymous
both plz!

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anonymous
  • anonymous
if you set them to zero just have 18m^2 -1=0 18m^2 =1 m^2 = 1/18 \[m = \sqrt{1/18}\]
anonymous
  • anonymous
so the same with the other you will have \[(m - \sqrt{1/18})^{2}\] and \[(m - \sqrt{1/81})^{2}\]
anonymous
  • anonymous
THANKZ YOUZ!
anonymous
  • anonymous
your welcome
asnaseer
  • asnaseer
@Davidjohn - I /think/ your last steps where you factorized the expressions has a mistake in it. I believe you can use the rule \(a^2-b^2=(a+b)(a-b)\) to get:\[81m^2-1=9^2m^2-1^2=(9m)^2-1^2=(9m+1)(9m-1)\]
anonymous
  • anonymous
@asnaseer that makes a whole lot more sense!
anonymous
  • anonymous
oh yeah, i generalized that one too much, youre right
anonymous
  • anonymous
but the answer would be right if expanded because it would simplify down and you could multiply them to get the same answer
asnaseer
  • asnaseer
glad I could help - and no worries @Davidjohn - we all make mistakes - that's what makes us human :-)
anonymous
  • anonymous
ah except it would be plus or minus then, not just minus minus
asnaseer
  • asnaseer
\[(m - \sqrt{1/81})^{2}=m^2-2\sqrt{1/81}+1/81\]
anonymous
  • anonymous
right my bad :) thanks for correcting
anonymous
  • anonymous
Woah... the first thing you showed me was a lot easier to understand...
asnaseer
  • asnaseer
:)

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