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Algebra Level 2

6 + 6 + 6 + ⋯ = ? \large \sqrt{6+\sqrt{6+\sqrt{6 +\cdots }}} =\, ?


The answer is 3.

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Nit Jon by Nit Jon , 21, USA

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1 solution

Nit Jon
Jan 30, 2016

First set this nested radical equal to a variable...

6 + 6 + 6 + . . . = x \sqrt{6+\sqrt{6+\sqrt{6 + ...}}} = x

( 6 + 6 + 6 + . . . ) 2 = x 2 (\sqrt{6+\sqrt{6+\sqrt{6 + ...}}})^2 = x^{2}

6 + 6 + 6 + . . . = x 2 6 + \sqrt{6+\sqrt{6 + ...}} = x^{2}

The tricky part is that you have to realize that the nested radical can be substituted with x x

6 + x = x 2 6 + x = x^{2}

Then you do old school quadratic...

x 2 − x − 6 = 0 x^{2}-x-6 = 0

We know that a square root can never be negative, so x x cannot be -2

x = 3 x = − 2 \boxed{x = 3} \quad x = -2

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