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.


Updated on: 10-Oct-2022

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