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(x-1)^2-13=3, how do you find the x?
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add 13 take the square root, don't forget the \(\pm\) then add 1
what?
add 13 to both sides what do you get?
16
actually you get \[(x-1)^2=16\]
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yes
can you think of a number whose square is 16?
4
ok so we can solve \(x-1=4\) easily enough now can you think of another number whose square is 16 (hint, it is negative)
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-4?
yes
so you have two equations to solve \[x-1=4\] and also \[x-1=-4\]
so will x be x=-5 and x=3
no
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to solve \[x-1=4\]you need to ADD 1 to both sides
5
and -3
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