Question

If x=1+√2, find the value of x^2 + 1/x^2

Answer 1

Answer:

5.999

Step-by-step explanation:

$x = 1+\sqrt{2} \\x= 2.414\\hence \\ 2.414^2 + \frac{1}{2.414^2} \\=5.828 + \frac{1}{5.828} \\= 5.999$

Answer 2

Heya!!!

SOLUTION ==>

$x = 1 + \sqrt{2}$

$Finding \: the \: value \: of \: \: \: \frac{1}{x}$

Putting the value of x

$\frac{1}{x} = \frac{1}{1 + \sqrt{2} }$

Rationalising

$\frac{1}{x} = \frac{1}{1 + \sqrt{2} } \times \frac{1 - \sqrt{2} }{1 - \sqrt{2} } \\ \\ \frac{1}{x} = \frac{1 - \sqrt{2} }{ {1}^{2} - ({ \sqrt{2} })^{2} } \\ \\ \frac{1}{x} = \frac{1 - \sqrt{2} }{1 - 2} \\ \\ \frac{1}{x} = \frac{1 - \sqrt{2} }{ - 1}$

$\frac{1}{x} = - (1 - \sqrt{2)} \\ \\ \frac{1}{x} = - 1 + \sqrt{2}$

Now,

$x + \frac{1}{x} = (1 + \sqrt{2} ) + ( - 1 + \sqrt{2)} \\ \\ x + \frac{1}{x} = 1 + \sqrt{2} - 1 + \sqrt{2} \\ \\ x + \frac{1}{x} = 2 \sqrt{2}$

On squaring both sides

${(x + \frac{1}{x} )}^{2} = {(2 \sqrt{2} )}^{2} \\ \\ {x}^{2} + { \frac{1}{x} }^{2} + 2 \times x \times \frac{1}{x} = 4 \times 2 \\ \\ {x}^{2} + { \frac{1}{x} }^{2} + 2 = 8 \\ \\ {x}^{2} + { \frac{1}{x} }^{2} = 8 - 2 \\ \\ {x}^{2} + { \frac{1}{x} }^{2} = 6$

Ans is 6

Hope it helps