Question

Evaluate :Underroot 20 +underroot 20 + underroot 20 +......

Answer 1

Let x = sqrt(20+sqrt(20+sqrt(20+.. and so on)

x = sqrt(20+x)

Squaring both sides,

x^2 = 20+x

x^2-x-20=0

x = 5 or x = -4 (not admissible as we can't get a square root of a number as negative)

Hence, the answer is 5

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