How do you solve #(y - 8) ^ { 2} = 49#?

2 Answers
Jun 25, 2017

#y=1" or " y=15#

Explanation:

#color(blue)"take the square root of both sides"#

#sqrt((y-8)^2)=+-sqrt49larrcolor(red)" note plus or minus"#

#rArry-8=+-7#

#"add 8 to both sides"#

#ycancel(-8)cancel(+8)=8+-7#

#rArry=8+-7#

#y=8+7rArry=15larr" is a solution"#

#y=8-7rArry=1larr" is a solution"#

#color(blue)"As a check"#

#y=15to(15-8)^2=7^2=49larr" true"#

#y=1to(1-8)^2=(-7)^2=49larr" true"#

Jun 25, 2017

15 or 1

Explanation:

Each side is a perfect square, so we just need to square root each side, so #(y-8)^2=49# becomes #y-8="+/-"7#. All numbers need to be put onto one side so #y-8="+/-"7# becomes #y=8"+/-"7#.

#y=8+7=15#
or
#y=-7+8=1#.