Square Roots Can be Big

Algebra Level 2

What is the value of x = ? x=? x = x 2 + 2020 − x 2 + 2020 − x 2 + 2020 − ⋯ x = \sqrt{x^2 + \sqrt{2020-\sqrt{x^2 + \sqrt{2020 - \sqrt{x^2 + \sqrt{2020 - \sqrt{\cdots}}}}}}}


The answer is 2020.

Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Chew-Seong Cheong
Dec 17, 2019

x = x 2 + 2020 − x 2 + 2020 − x 2 + 2020 − ⋯ x = x 2 + 2020 − x 2 + 2020 − x Squaring both sides x 2 = x 2 + 2020 − x ⟹ x = 2020 \begin{aligned} x & = \sqrt{x^2 + \sqrt{2020-\sqrt{x^2 + \sqrt{2020 - \sqrt{x^2 + 2020 - \sqrt{\cdots}}}}}} \\ x & = \sqrt{x^2 + \sqrt{2020-\sqrt{x^2 + \sqrt{2020 - x}}}} & \small \blue{\text{Squaring both sides}} \\ x^2 & = x^2 + 2020 - x \\ \implies x & = \boxed {2020} \end{aligned}

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Anurag Biswas
Dec 17, 2019

x 2 = x 2 + 2020 − x 2 + 2020 − … x^2 = x^2 +\sqrt{2020-\sqrt{x^2+\sqrt{2020 - \ldots}}}

⇒ 2020 − x 2 + 2020 − … = 0 \Rightarrow\sqrt{2020-\sqrt{x^2+\sqrt{2020 - \ldots}}} = 0

⇒ x 2 + 2020 − … = 2020 \Rightarrow \sqrt{x^2+\sqrt{2020 - \ldots}} = 2020

S i n c e x = x 2 + 2020 − x 2 + 2020 − … Since\,x = \sqrt{x^2 +\sqrt{2020-\sqrt{x^2+\sqrt{2020 - \ldots}}}}

S o x = 2020 So \,x = 2020

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