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:

$x = +-11$

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