How do you solve #1/4(b-8)^2=7#?

1 Answer
Apr 14, 2017

Answer:

#b=8+-2sqrt7#

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"#