Question

If P(x) = x2-2V2x + 1, then find P(272) !
ch do not lie in any of the quadrants? If
$if$
$if \: p(x) = x2 - 2 \sqrt{2x} + 1 \: then \: find(2 \sqrt{2} )$
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Answer 1

Answer:

p(2√2)= (2√2)^2-2√2×2√2+1

p(2√2)= 1