How do you solve #(5x+1)^2=25#?

Question

2749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-07T00:35:15

#x = 4/5# and #x = -6/5#

#sqrt((5x + 1)^2) = sqrt(25)#

#:.5x+1=+-5#

#5x + 1 = 5# and #5x + 1 = -5#

#5x + 1 - 1 = 5 - 1#

#5x = 4#

#(5x)/5 = 4/5#

#x = 4/5#

and

#5x + 1 - 1 = -5 - 1#

#5x = -6#

#(5x)/5 = -6/5#

#x = -6/5#