How do you solve #(x-9)^2=12#?

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Answer 1 · 2018-03-30T02:39:47

#x=-2sqrt3+9, x=-2sqrt3+9#

You could expand the left side, move everything to the left, and apply the quadratic formula; however, taking the root of both sides is far faster:

#sqrt((x-9)^2)=+-sqrt(12)#

Recall that #sqrt(a^2)=a#, and that #sqrt12=sqrt(4*3)=sqrt4sqrt3=2sqrt3# :

#x-9=+-2sqrt3#

#x=-2sqrt3+9, x=-2sqrt3+9#

You would have the same result even with applying the quadratic formula -- this method is just faster and cleaner.

Answer 2 · 2018-03-30T02:44:14

#x=2sqrt3+9,# #-2sqrt3+9#

Solve:

#(x-9)^2=12#

Take the square root of both sides.

#x-9=+-sqrt12#

Prime factorize #12#.

#x-9=+-sqrt(2xx2xx3)=#

#x-9=+-sqrt(2^2xx3)#

Apply rule: #sqrt(a^2)=a#

#x-9=+-2sqrt3#

Add #9# to both sides.

#x=+-2sqrt3+9#

Solutions for #x#.

#x=2sqrt3+9,# #-2sqrt3+9#