Question:
If $p(x)=x^{2}-2 \sqrt{2} x+1$, then $p(2 \sqrt{2})=$ ?
(a) 0
(b) 1
(c) $4 \sqrt{2}$
(d) $-1$
Solution:
(b) 1
$p(x)=x^{2}-2 \sqrt{2} x+1$
$\therefore p(2 \sqrt{2})=(2 \sqrt{2})^{2}-2 \sqrt{2} \times(2 \sqrt{2})+1$
$=8-8+1$
$=1$
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