How do you solve $x^{2} - 121$ $x^2 - 121$ ?

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Answer 1 · 2018-04-01T20:33:52

$x = 11, x = - 11$ $x=11, x=-11$

Using the difference of squares, which states that

$(x^{2} - a^{2}) = (x + a) (x - a)$ $(x^2-a^2)=(x+a)(x-a)$ :

$(x^{2} - 121) = 0$ $(x^2-121)=0$

$(x + 11) (x - 11) = 0$ $(x+11)(x-11)=0$

$x = 11, x = - 11$ $x=11, x=-11$

How do you solve x2−121x^2 - 121?