Nested Radicals (2)
Rad Chad, after seeing the first problem , wanted to know if he could generalize it. So, he starts with the equation x = k + k + k + ⋯ . Then x 2 − k ⇒ x 2 − x − k = k + k + k + … = x = 0 . Chad then used the quadratic formula to solve for x as a function of k ( with a = 1 , b = − 1 , c = − k ) : x = 2 ( 1 ) − ( − 1 ) + ( − 1 ) 2 − 4 ( 1 ) ( − k ) = 2 1 + 1 + 4 k . Being the good mathematician he is, Chad checks his model with the previous example: 2 1 + 1 + 4 ( 1 2 ) = 2 8 = 4 . Is Chad's model correct for all values of k ?
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1 solution
Why is it true when c=12 ?
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It is true for most numbers, but not for all. I will answer your question in a later problem.
This explains why "Yes" is not the correct answer.
Why is the given answer correct?
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The lack of a negative symbol at the beginning implies that ± is not necessary. It would be nonsense to say that 1 2 + 1 2 + 1 2 + … is equal to -3.
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Is it not equivalent to say 0 + 0 + 0 + ⋯ = 2 1 + 1 + 4 ( 0 ) is nonsense, but that 0 + 0 + 0 + ⋯ = 2 1 − 1 + 4 ( 0 ) is true, so by this example, we must have c + c + c + ⋯ = 2 1 − 1 + 4 c ?
Of course, I'm being a little facetious, but your response was a bit of a non sequitur. I asked why the "correct" answer is correct, not why we could ignore the minus sign for c > 0 (which is really the furthest I can extend your response)
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@Brian Moehring – I do see your point. Zero is always a very strange number. Feel free to leave a report and tag a moderator, or see the discussion thread .
For c=0, the answer would be 1, which is incorrect.