If $p(x)=x^{2}-2 \sqrt{2} x+1$, then what is the value of $p(2 \sqrt{2})$
Given :
The given expression is $p(x)=x^{2}-2 \sqrt{2} x+1$.
To find :
We have to find the value of $p(2 \sqrt{2})$.
Solution :
$p(x)=x^{2}-2 \sqrt{2} x+1$
Substitute $x = 2\sqrt{2}$
$p(2 \sqrt{2}) = (2 \sqrt{2})^{2}-2 \sqrt{2}(2 \sqrt{2})+1$
$ = 4 \times 2 - 4 \times 2 + 1$
$ = 8 - 8 + 1 = 0 + 1$
$ = 1.$
The value of $p(x)=x^{2}-2 \sqrt{2} x+1$ at $p(2 \sqrt{2})$ is 1.
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