Question

if x=√2+1,find the value of [x-1/x]^2​

Answer 1

Step-by-step explanation:

Value of x^2+\frac{1}{x^2}x

2

+

x

2

1

is 6.

Step-by-step explanation:

Given: x=1+\sqrt{2}x=1+

2

To find: x^2+\frac{1}{x^2}x

2

+

x

2

1

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}}

x

2

1

=

3+2

2

1

=

3+2

2

1

×

3−2

2

3−2

2

=\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2})^2}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}=

3

2

−(2

2

)

2

3−2

2

=

9−8

3−2

2

=3−2

2

Now,

x^2+\frac{1}{x^2}=3+2\sqrt{2}+3-2\sqrt{2}=6x

2

+

x

2

1

=3+2

2

+3−2

2

=6

Therefore, Value of x^2+\frac{1}{x^2}x

2

+

x

2

1

is 6