Root 11 to the infinity

Algebra Level 2

$\large \sqrt{\color{#624F41}{11} + \sqrt {\color{#302B94}{11} + \sqrt{\color{#20A900}{11} + \sqrt {\color{#EC7300}{11} + \sqrt {\color{#E81990}{11} \ldots }}}}}$

What is the value of the expression above? Round your answer off to nearest whole number.

4 3 2 0 11

1 solution

A Former Brilliant Member
Nov 4, 2015

$\large \sqrt{\color{#624F41}{11} + \sqrt {\color{#302B94}{11} + \sqrt{\color{#20A900}{11} + \sqrt {\color{#EC7300}{11} + \sqrt {\color{#E81990}{11} \ldots }}}}}=x$

Squaring both sides:

$11 \large \sqrt {\color{#302B94}{11} + \sqrt{\color{#20A900}{11} + \sqrt {\color{#EC7300}{11} + \sqrt {\color{#E81990}{11} \ldots}}}}=x^2$

Take the $11$ onto the other side of the equation:

$\ \large \sqrt {\color{#302B94}{11} + \sqrt{\color{#20A900}{11} + \sqrt {\color{#EC7300}{11} + \sqrt {\color{#E81990}{11} \ldots }}}}=x^2-11$

And then we remember that

$\large \sqrt{\color{#624F41}{11} + \sqrt {\color{#302B94}{11} + \sqrt{\color{#20A900}{11} + \sqrt {\color{#EC7300}{11} + \sqrt {\color{#E81990}{11} \ldots }}}}}=x$ so

$x^2-x-11=0$ .

Complete the square. The positive solution is $\approx 3.85$ (done by hand), so the answer is $4$ to the nearest whole number.

The possible answers should be made more difficult... Clearly the answer must be greater than 3 since the radicand is increasing from a number greater than 9. Yet 11 is much too large as 121 could never be reached. Andrei's method is elegant (and of course preferred), but a simple educated guess works just fine since answers don't include the 5-7 range.

Mark Wo - 5 years, 7 months ago

Root 11 to the infinity

1 solution

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