How do you solve # (x-5)^2=4#?

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Answer 1 · 2017-06-15T12:13:39

Solution: #x=7 or x=3#

#(x-5)^2=4 or x-5 =sqrt4 or x-5 = +-2 or x=5+-2 :. x=7 or x=3#

Solution: #x=7 or x=3# [Ans]

Answer 2 · 2017-06-15T12:15:44

#x=3or x=7#

If #(x-5)^2=4#
then
#color(white)("XXX")x-5=+-sqrt(4)=+-2#

#{:("if " x-5=-2,color(white)("xxx"),"if "x=+2), ("then " x=3,,"then "x=7) :}#