How do you solve #(x + 2)^2 = 16#?

3 Answers

Answer:

#x=2, -6#

Explanation:

#(x+2)^2=(+-4)^2#
#sqrt((x+2)^2)=sqrt (+-4^2)#
#x+2=+-4#
#x=2, -6#

Mar 8, 2016

Answer:

#x=2 or x=-6#

Explanation:

#(x+2)^2=16#

#x+2=+-4#

#x=2 or x=-6#

Answer:

#x=2#

Explanation:

#(x+2)^2=16#

Take the Square root of both sides

#rarrsqrt((x+2)^2)=sqrt16#

#rarrx+2=+-4#

#rarrx=+-4-2#

#rArrx=2, -6#

Check

#(2+2)^2=16#

#(4)^2=16#

#16=16# =)

~~~

#(2-6)^2=16#

#(-4)^2=16#

#16=16#