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Concept of Square Root

If x=1+√2, ...

Question

If $x = 1 + \sqrt{2}$ , then $x^{2} - 2 x + 1$ equals

A

1 + \sqrt{2}

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B

\sqrt{2} - 1

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C

2

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D

\sqrt{2}

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E

2 + \sqrt{2}

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Solution

The correct option is D $2$
Given $x = 1 + \sqrt{2}$

$\Rightarrow x^{2} - 2 x + 1 = 3 + 2 \sqrt{2} - 2 - 2 \sqrt{2} + 1 = 2$

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Similar questions

x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}, y = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}

x^{2} + x y + y^{2}

I = \int \frac{d x}{1 + \sqrt{x^{2} + 2 x + 2}} = \frac{A}{7} log ∣ ∣ x + 1 + \sqrt{x^{2} + 2 x + 2} ∣ ∣ - \frac{\sqrt{x^{2} + 2 x + 2} - 1}{x + 1} + C

then A is equal to

s i n^{- 1} \sqrt{x^{2} + 2 x + 1} + s e c^{- 1} \sqrt{x^{2} + 2 x + 1} = \frac{π}{2}, x \neq 0

then the value of

2 s e c^{- 1} (\frac{x}{2}) + s i n^{- 1} (\frac{x}{2})

f (x) = s i n^{- 1} (\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}), - \frac{1}{2} \leq x \leq 1

f (x)

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

, then x is equal to:

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