If $p(x) = x^2 - 2\sqrt{2}x+1$, then find the value of $p(2\sqrt{2})$.
Given :
The given expression is $p(x) = x^2 - 2\sqrt{2}x+1$.
To do :
We have to find the value of $p(2\sqrt{2})$.
Solution :
$p(x) = x^2 - 2\sqrt{2}x+1$
$p(2\sqrt{2}) = (2\sqrt{2})^2 - 2\sqrt{2} (2\sqrt{2}) +1$
$= (2\sqrt{2})^2 -(2\sqrt{2})^2 +1 $
$ = 0 + 1 = 1$
Therefore, the value of $p(2\sqrt{2})$ is 1.
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