How do you solve ${(x - 9)}^{2} = 1$ $(x - 9)^2 = 1$ ?

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Answer 1 · 2016-07-30T18:59:15

If a number to the square is $1$ $1$ , that number is either $1$ $1$ or $- 1$ $-1$

Thus if ${(x - 9)}^{2} = 1$ $(x-9)^2=1$ we know that either $(x - 9) = 1$ $(x-9)=1$ or $(x - 9) = - 1$ $(x-9)=-1$ .

Hence we have either $x = 10$ $x=10$ or $x = 8$ $x=8$

How do you solve (x−9)2=1(x - 9)^2 = 1?