Question

If $p\left(x\right)=x^2-2\sqrt{2}x+1$, then $p\left(2\sqrt{2}\right)$ is equal to

A) $0$
B) $1$
C) $4\sqrt{2}$
D) $8\sqrt{2}+1$

Answer 1

$p\left(x\right)=x^2-2\sqrt{2}x+1$
On putting $x=2\sqrt{2}$

$p\left(2\sqrt{2}\right)=(2\sqrt{2}{)}^2-(2\sqrt{2}\left)\right(2\sqrt{2})+1$

$=8-8+1=1$

$p\left(2\sqrt{2}\right)=1$

So, the answer is option B.