If \[p\left( x \right) = {x^2} - 2\sqrt {2x} + 1\]then \[p\left( {2\sqrt 2 } \right)\] is equal to
Answer
486.6k+ views
We have
\[\begin{gathered}
p\left( x \right) = {x^2} - 2\sqrt {2x} + 1 \\
p\left( {2\sqrt 2 } \right){\left( {2\sqrt 2 } \right)^2} - 2\sqrt 2 \left( {2\sqrt 2 } \right) + 1 \\
= 8 - 8 + 1 \\
= 1 \\
\end{gathered} \]
Hence, (b) is the correct answer.
\[\begin{gathered}
p\left( x \right) = {x^2} - 2\sqrt {2x} + 1 \\
p\left( {2\sqrt 2 } \right){\left( {2\sqrt 2 } \right)^2} - 2\sqrt 2 \left( {2\sqrt 2 } \right) + 1 \\
= 8 - 8 + 1 \\
= 1 \\
\end{gathered} \]
Hence, (b) is the correct answer.
Last updated date: 28th Sep 2023
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