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Value of an infinite surdWhat is the value of $\displaystyle \sqrt{3 + 2\sqrt{3 + 2\sqrt{3 + 2\sqrt{3 + \cdots}}}}$? What is a general way of finding the value of infinite surds like this? |

If is has a value, $s$, then we can take your expression, multiply by 2, add 3 and take the square root and be left with the same value. That is, $s = \sqrt{3+2s}$ Alternatively, if you square it, subtract 3 and divide by 2, you are left with the same value: $s=\frac12(s^2-3)$. (Note that this gives, in addition to the positive solution, a negative solution which in an obvious - but not necessarily correct - way cannot be valid). I get $s \in \{-1,3\}$. In other words, the general way of solving these is to see what operations are required to add or remove a "term". |

I think I would go so far as to say that the negative solution is not correct because the square root function is never negative for finite input. The argument can be made more technical and less certain, but in the end it boils down to the above. |

first we should claim if the infinite surd converges before we can equal it as a real number |

That was the very first thing that I said. Except that, strictly speaking, this is not a limit problem so there is no "convergence". My phraseology reflects that. If the question talked about limits or convergence, you would form a series based on the recurrence formula $s_{n+1}=\sqrt{3+2s_n}$ and show that it converges for all positive $s_1$. Actually, I think it converges to 3 for all $s_1 \ge -\frac32$. For other $s_1$, $s_2$ is not real. |

Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers. |

Quote:
show some effort. |

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