How do you solve ${(x - 3)}^{2} = 16$ $(x-3)^2=16$ ?

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Answer 1 · 2016-08-23T10:21:39

$x = 7$ $x=7$
$x = - 1$ $x=-1$

${(x - 3)}^{2} = 16$ $(x-3)^2=16$

or

$x - 3 = \sqrt{16}$ $x-3=sqrt16$

or

$x - 3 = \pm 4$ $x-3=+-4$

or

$x = 3 \pm 4$ $x=3+-4$

or

$x = 3 + 4$ $x=3+4$

or

$x = 7$ $x=7$ ----------------------Ans $1$ $1$

or

$x = 3 - 4$ $x=3-4$

or

$x = - 1$ $x=-1$ -------------------------Ans $2$ $2$

How do you solve (x-3)^2=16(x−3)2=16(x-3)^2=16?