The value of -2-2- -2- - - - is -

Let x = √(2+√(2+ √(2+ ........... ))) nbsp; By squaring on both sides, we get ⇒ x^2 = 2 + nbsp;√(2+√(2+ √(2+ ... ∞))) ⇒ x^2 = 2 + x ⇒ x^2 x 2 = 0 ⇒(x ...

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Byju's Answer

Standard X

Mathematics

Solving Using Quadratic Formula When D>0

The value of ...

Question

The value of $\sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . . . . .}}}$ is _______.

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Solution

The correct option is A 2
Let $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . . . . . . .}}}$
By squaring on both sides, we get

$\Rightarrow x^{2} = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + . . . \infty}}} \Rightarrow x^{2} = 2 + x \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$

We know, $\sqrt{2} = 1.41$ so, negative value to be neglected.

$∴ x = 2 \Rightarrow \sqrt{2 + \sqrt{2 + \sqrt{2 + . . .}}} = 2$

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The value of √2+√2+√2+... ... ... is _______.

The correct option is A 2Let x=√2+√2+√2+........... By squaring on both sides, we get ⇒x2=2+√2+√2+√2+...∞⇒x2=2+x⇒x2−x−2=0⇒(x−2)(x+1)=0⇒x=−1,2 We know, √2=1.41 so, negative value to be neglected. ∴x=2⇒√2+√2+√2+...=2

The value of $\sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . . . . .}}}$ is _______.