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

Question

3503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-25T07:09:24

#y=0# or #y=-5#

writing
#(2y+5)^2-5^2=0#
#(2y+5-5)(2y+5+5)=0#
this gives
#2y=0#
or
#2y+10=0#
this gives
#y=0# or #y=-5#

Answer 2 · 2018-05-25T07:46:47

#y=0,-5#

Solve:

#(2y+5)^2=25#

Take the square root of both sides.

#sqrt((2y+5)^2)=sqrt25#

#2y+5=+-5#

This gives us two equations:

#color(blue)(2y+5=+5# and #color(green)(2y+5=-5#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(2y+5=5#

Subtract #5# from both sides.

#2y+5-5=+5-5#

#2y=0#

Divide both sides by #2#.

#y=0/2#

#y=0#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(green)(2y+5=-5#

Subtract #5# from both sides.

#2y+5-5=-5-5#

#2y=-5-5#

#2y=-10#

Divide both sides by #2#.

#y=-10/2#

#y=-5#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Solutions for #y#.

#y=0,-5#