Question

x= √2+1, find x²+1/x²​

Answer 1

Answer:

Value of x^2+\frac{1}{x^2}x2+x21 is 6.

Step-by-step explanation:

Given: x=1+\sqrt{2}x=1+2

To find: x^2+\frac{1}{x^2}x2+x21

x² = ( 1 + √2 )² = 1² + (√2)² + 2 × 1 × √2 = 1 + 2 + 2√2 = 3 + 2√2

\frac{1}{x^2}=\frac{1}{3+2\sqrt{2}}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}x21=3+221=3+221×3−223−22

                        =\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2})^2}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}=32−(22)23−22=9−83−22=3−22

Now,

x^2+\frac{1}{x^2}=3+2\sqrt{2}+3-2\sqrt{2}=6x2+x21=3+22+3−22=6

Therefore, Value of x^2+\frac{1}{x^2}x2+x21 is 6