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Mathematics 7 Online
OpenStudy (anonymous):

18m^2-1 factor plz!

OpenStudy (anonymous):

\[81m^2-1\]

OpenStudy (anonymous):

wait which one, or both

OpenStudy (anonymous):

both plz!

OpenStudy (anonymous):

if you set them to zero just have 18m^2 -1=0 18m^2 =1 m^2 = 1/18 \[m = \sqrt{1/18}\]

OpenStudy (anonymous):

so the same with the other you will have \[(m - \sqrt{1/18})^{2}\] and \[(m - \sqrt{1/81})^{2}\]

OpenStudy (anonymous):

THANKZ YOUZ!

OpenStudy (anonymous):

your welcome

OpenStudy (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:\[81m^2-1=9^2m^2-1^2=(9m)^2-1^2=(9m+1)(9m-1)\]

OpenStudy (anonymous):

@asnaseer that makes a whole lot more sense!

OpenStudy (anonymous):

oh yeah, i generalized that one too much, youre right

OpenStudy (anonymous):

but the answer would be right if expanded because it would simplify down and you could multiply them to get the same answer

OpenStudy (asnaseer):

glad I could help - and no worries @Davidjohn - we all make mistakes - that's what makes us human :-)

OpenStudy (anonymous):

ah except it would be plus or minus then, not just minus minus

OpenStudy (asnaseer):

\[(m - \sqrt{1/81})^{2}=m^2-2\sqrt{1/81}+1/81\]

OpenStudy (anonymous):

right my bad :) thanks for correcting

OpenStudy (anonymous):

Woah... the first thing you showed me was a lot easier to understand...

OpenStudy (asnaseer):

:)

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