How do you solve $(p-4)^2=16$ ?

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Answer 1 · 2016-08-31T21:27:47

$p = 8 or p = 0$

Although this is a quadratic, it is a special case in that there is no term in $x$ .

We can simply find the square root of each side,

$sqrt((p-4)^2) = +-sqrt16$

$p-4 = +-4$

$p-4 = +4 rarr p = 8$

$p-4 =-4 rarr p = 0$

Answer 2 · 2016-08-31T21:30:52

P=0 or 8

$(p-4)^2=16$
Take square root of both sides
$(p-4)=+-4$
So p-4=-4 which gives p=0

Or
p-4=4 which gives p=8

How do you solve (p-4)^2=16(p-4)^2=16?