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

1 Answer
Jim G.
Apr 14, 2017

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"

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