How do you solve #1/4(b-8)^2=7#?
1 Answer
Apr 14, 2017
Explanation:
#"multiply both sides of the equation by 4"#
#cancel(4)^1xx1/cancel(4)^1(b-8)^2=4xx7#
#rArr(b-8)^2=28#
#color(blue)"take the square root of both sides"#
#sqrt((b-8)^2)=+-sqrt28#
#rArrb-8=+-sqrt(4xx7)# add 8 to both sides.
#bcancel(-8)cancel(+8)=8+-2sqrt7#
#rArrb=8+-2sqrt7" are the solutions"#
