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anonymous
 5 years ago
I need help. I don't understand how to solve this equation.
z^29z+18=0
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anonymous
 5 years ago
I need help. I don't understand how to solve this equation. z^29z+18=0 Show Steps.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first factor it (z6)(z3)=0 z = 6 or 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because either z6 = 0 or z3 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(b(+)sqrt(b^24ac))/2a. a=w^2 = 1 b=7w = 7 c=8 w= (7+sqrt(494*(1*8)))/2 = (7+sqrt(81))/2 = (7+9)/2 = 16/2 and 2/2 => w=8 V w=1 I wrote this for a similar question. Just insert for a,b and c, and z instead of w, from the equation you provided. V is mathematical notation, meaning "and". So you'll have two answers for z, both equally correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u don't need to do this as the function factorises

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks jimmyrep, I don't understand Msc.student ... Im only in Algebra 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, but factorisation requires some intuition, meaning you'll have to have solved a lot of problems using it, to get the answers right consistently. I think a better place to start is the second degree equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not familiar with how the American school works, so Algebra 1 doesn't mean much to me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, okay. Factorisation it is, then. You'll become familiar with the second degree equation soon enough.
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