Question 364929
(k - 1) / 2 = [sqrt](k - 1)
k -1 = 2 [sqrt](k - 1)
Square both sides of the equation
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k^2 - 2k + 1 = 4 (k - 1)
k^2 - 2k + 1 = 4k - 4
k^2 - 6k + 5 = 0
(k - 1) (k - 5) = 0
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k - 1 = 0
k = 1
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k - 5 =0
k = 5
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So the values of k are 1 and 5.