How do you solve #(p - 4) ^ { 2} = 49#?

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Answer 1 · 2016-10-25T19:42:15

#p = 11# or #p = -3#

First, take the Square Root of each side of the equation:

#sqrt((p - 4) ^ 2) = sqrt(49)#

Which leaves:

#p - 4 = 7# or #p - 4 = -7#

Solving for #p# gives:

#p - 4 + 4 = 7 + 4#

#p = 11#

or

#p - 4 + 4 = -7 + 4#

#p = -3#

Answer 2 · 2016-10-25T19:44:14

#p=11 or p=-3#

#(p-4)^2=49#
#(p-4)^2=(7)^2#

We can now square root both sides

#sqrt((p-4)^2)=sqrt((7)^2)#
#p-4=+-7#

If 7 is positive
#p=7+4#
#p=11#

If 7 is negative
#p=-7+4#
#p=-3#