How do you solve #(p-4)^2=16#?

Question

4919 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T21:27:47

#p = 8 or p = 0#

Although this is a quadratic, it is a special case in that there is no term in #x#.

We can simply find the square root of each side,

#sqrt((p-4)^2) = +-sqrt16#

#p-4 = +-4#

#p-4 = +4 rarr p = 8#

#p-4 =-4 rarr p = 0#

Answer 2 · 2016-08-31T21:30:52

P=0 or 8

#(p-4)^2=16#
Take square root of both sides
#(p-4)=+-4#
So p-4=-4 which gives p=0

Or
p-4=4 which gives p=8