If p(x) = x^{2} - 2sqrt{2}x + 1,then p(2sqrt{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the polynomial $p(x) = x^{2} - 2\sqrt{2}x + 1$.To find $p(2\sqrt{2})$,we substitute $x = 2\sqrt{2}$ into the polynomial expression.$p(2\sqrt

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the polynomial $p(x) = x^{2} - 2\sqrt{2}x + 1$.To find $p(2\sqrt{2})$,we substitute $x = 2\sqrt{2}$ into the polynomial expression.$p(2\sqrt{2}) = (2\sqrt{2})^{2} - 2\sqrt{2}(2\sqrt{2}) + 1$Calculating the terms:$(2\sqrt{2})^{2} = 4 	imes 2 = 8$$2\sqrt{2}(2\sqrt{2}) = 4 	imes 2 = 8$Substituting these back into the expression:$p(2\sqrt{2}) = 8 - 8 + 1$$p(2\sqrt{2}) = 1$Therefore,the correct option is $(b)$.

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

Similar Questions

Factorise : $2x^{3}-3x^{2}-17x+30$

Verify whether the following is True or False:
$-3$ is a zero of $x-3$.

Find the value of $x^{3}+y^{3}-12xy+64$,when $x+y=-4$.

Write the degree of the following polynomial:
$11-2 y^{2}$

The value of the polynomial $5x - 4x^2 + 3$ when $x = -1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module