root2 + root 2 + root 2 + root 2 + root 2
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Answer:
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+.....................\infty }}}}
Let,
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+.....................\infty }}}}=x
\sqrt{2+x}=x
2+x=x^2
\implies x^2 - x - 2 =0
x^2 - 2x + x - 2=0
x(x-2)+1(x-2)=0
(x+1)(x-2)=0
⇒ x = - 1 or x = 2,
x = - 1 is not possible, because value of square root of a number can not be negative,
\implies \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+.....................\infty }}}} = 2
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Answer:
5 root 2
Step-by-step explanation:
5 root 2
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