If x - -2 - -2 - -2 - - then

The correct option is D x = 2 x=√(2 + √(2 + √(2 + ...))) x = √(2 + x) x^2 = x + 2 ⇒ x^2 x 2=0 ⇒ (x 2)(x+1)=0 ⇒ x = 1, 2 But ...

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Relation between Square and Square Root

If x = √2 +...

Question

If $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + . . .}}}$ then

x = - 1

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- 1 < x < 2

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x = 2

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x = 3

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Solution

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Similar questions

x^{4} + x^{3} - 2 x^{2} + x + 1

= x - 1

= 2

$If the dividend = x^{4} + x^{3} - 2 x^{2} + x + 1, divisor = x - 1 and remainder = 2, then find the quotient q(x) .$

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 (x - 1) & 3 (x - 1) (x - 2) x - 1 & (x - 1) (x - 2) & (x - 1) (x - 2) (x - 3) x & (x - 1) x & (x - 1) (x - 2) x \end{matrix} ∣ ∣ ∣ ∣ ∣

f (41)

(x - 1)^{2} (x - 2)^{2}

(x - 1) (x - 2)

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

α =

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