How do you solve #x^2 - 121 = 0#?

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Answer 1 · 2018-05-21T06:07:04

#x=11" "# or #" "x=-11#

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

Note that both #x^2# and #121=11^2# are perfect squares.

So we find:

#0 = x^2-121 = x^2-11^2 = (x-11)(x+11)#

So:

#x = +-11#