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

If a number to the square is $1$, that number is either $1$ or $- 1$
Thus if ${\left(x - 9\right)}^{2} = 1$ we know that either $\left(x - 9\right) = 1$ or $\left(x - 9\right) = - 1$.
Hence we have either $x = 10$ or $x = 8$