Math, asked by swanhayden7, 7 days ago

solve the question

 \sqrt{2 +  \sqrt{2}  +  \sqrt{2 +  \sqrt{2 +  ... \infty } } }

Answers

Answered by mathdude500
9

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\: \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  -  -  -  \infty } } } }

Let assume that

\rm :\longmapsto\: x = \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  -  -  -  \infty } } } }

On squaring both sides, we get

\rm :\longmapsto\:  {x}^{2} = 2 +  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  -  -  -  \infty } } } }

\rm :\longmapsto\:  {x}^{2} = 2 +  x

\rm :\longmapsto\:  {x}^{2} - 2  - x = 0

\rm :\longmapsto\:  {x}^{2} - x  - 2 = 0

\rm :\longmapsto\:  {x}^{2} - 2x + x  - 2 = 0

\rm :\longmapsto\:x(x - 2) + 1(x - 2) = 0

\rm :\longmapsto\:(x - 2)(x+ 1)= 0

\rm\implies \:x = 2 \:  \: or \:  \: x =  - 1

\rm :\longmapsto\: As \: x = \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  -  -  -  \infty } } } } > 0

Thus,

\bf\implies \:x = 2

Hence,

\rm\implies \:\boxed{\tt{  \sqrt{2 +  \sqrt{2 +  \sqrt{2 +  -  -  -  \infty } } }  = 2}}

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

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➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)

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