How do you solve #( x + 1 )^2 =25#?

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Answer 1 · 2016-05-08T18:31:31

Use the difference of squares identity to find:

#x=4# or #x=-6#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(x+1)# and #b=5# as follows...

Subtract #25# from both sides of the equation to find:

#0 = (x+1)^2-25#

#= (x+1)^2-5^2#

#=((x+1)-5)((x+1)+5)#

#=(x-4)(x+6)#

Hence #x=4# or #x=-6#