If $x = \frac{1}{(\sqrt{2} + 1)},$ $x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{,}}$ the value of x² + 2x - 1 is?

1 $2 \sqrt{2}$ ${\text{2}}\sqrt 2$
24
30
42

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given that, \\ x = \frac{1}{(\sqrt{2} + 1)} \\ Then, x^{2} + 2 x - 1 \\ = x^{2} + 2 x - 1 + 1 - 1 \\ = x^{2} + 2 x + 1 - 2 \\ = {(x + 1)}^{2} - 2 \\ Now put the value of x \\ = {(\frac{1}{(\sqrt{2} + 1)} + 1)}^{2} - 2 \\ = {(\frac{1 + \sqrt{2} + 1}{\sqrt{2} + 1})}^{2} - 2 \\ = {(\frac{\sqrt{2} + 2}{\sqrt{2} + 1})}^{2} - 2 \\ = {(\frac{(\sqrt{2} + 1) \times 2}{\sqrt{2} + 1})}^{2} - 2 \\ = {(\sqrt{2})}^{2} - 2 \\ = 2 - 2 \\ = 0 \end{aligned}

$\eqalign{ & {\text{Given that,}} \cr & {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr & {\text{Then, }}{x^2} + 2x - 1 \cr & = {x^2} + 2x - 1 + 1 - 1 \cr & = {x^2} + 2x + 1 - 2 \cr & = {\left( {x + 1} \right)^2} - 2 \cr & {\text{Now put the value of }}x \cr & = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr & = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\sqrt 2 } \right)^2} - 2 \cr & = 2 - 2 \cr & = 0 \cr}$