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anonymous
 5 years ago
can anyone give me some examples of problems like " whats is the solution set for (x=1)^2=49?
anonymous
 5 years ago
can anyone give me some examples of problems like " whats is the solution set for (x=1)^2=49?

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radar
 5 years ago
Best ResponseYou've already chosen the best response.0Have you stated the example correctly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dang it. no...it was (x+1)^2=49

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x+1)^2=7^2 so (x+1)=plus minus (7) x=plus minus 6

radar
 5 years ago
Best ResponseYou've already chosen the best response.0O.K look like 6 would do it so would 8. Lets work it out. \[(x+1)^{2}= x ^{2}+2x+1=49\] \[x ^{2}+2x48=0\] \[(x+8)(x6)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm..the answer choices are in format of ( 48,48) and that kind of thing

radar
 5 years ago
Best ResponseYou've already chosen the best response.0(x6)=0, x=6 (x+8)=0, x=8

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The solution set is x= 6, 8

radar
 5 years ago
Best ResponseYou've already chosen the best response.0What exactly is asked for the example?

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Are you working on the "completing the square" method, or factoring or quadratic? for solutions??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just solution sets. its part of the algebra 2 EOC, so i want to learn how to work them out

radar
 5 years ago
Best ResponseYou've already chosen the best response.0method I used is called the "factoring method" The quadratic equation was manipulated to equal 0 and then factored. (x+8)(x6) =0 The reason this works, one or both of the factors has to equal 0 for the results to equal 0. Like, ab=0, then a or b or both equal 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0um...im actually not that good at math, just found this site..wish i had found it sooner though. i could have asked for help on my last project
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