Question

If p(y) =y2

- 3√2y + 1 then find p(3√2)​

Answer 1

$Given \: p(y) = y^{2} - 3\sqrt{2} y + 1$

$Value \: of \: p(3\sqrt{2}) = (3\sqrt{2})^{2} - (3\sqrt{2})(3\sqrt{2}) + 1$

$Substitute\: y = (3\sqrt{2}) \: in \: the \: equation, \\we \:get$

$= \cancel {(3\sqrt{2})^{2} } - \cancel {(3\sqrt{2})^{2} }+ 1$

$= 1$

Therefore.,

$\red{Value \: of \: p(3\sqrt{2})}\green { = 1 }$

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Answer 2

Answer:

Hope it helps a lot .,..........