Question #256cc

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Answer 1 · 2016-10-26T07:07:24

#k = (3+sqrt(9+8N))/2#

#N = (k^2-3k)/2#

#=> 2N = k^2-3k#

#=> k^2-3k-2N = 0#

We can now apply the quadratic formula with #a=1, b = -3, c = -2N#:

#k = (-(-3)+-sqrt((-3)^2-4(1)(-2N)))/(2(1))#

#=> k = (3+-sqrt(9+8N))/2#

Normally, we would be done. However, we are given that #k>0# and #N>0#. As #N>0#, we have

#N>0#

#=> sqrt(9+8N) > sqrt(9) = 3#

#=> 3 - sqrt(9+8N) < 0#

#=> (3-sqrt(9+8N))/2 < 0#

As we only want a positive value for #k#, we will dismiss the negative result, leaving us with a single answer:

#k = (3+sqrt(9+8N))/2#