Problem:
How many distinct values of x satisfy ⌊x⌋2−3x+2=0, where ⌊x⌋ denotes the greatest integer less than or equal to x?
Answer Choices:
A. an infinite number
B. 4
C. 2
D. 3
E. 0
Solution:
First consider the possibility of integer solutions x=k. Then k2−3k+2=(k−2)(k−1)=0 has solutions k=1 and k=2.\
Next consider the possibility of a noninteger solution x where k<x<k+1 for some integer k. Then ⌊x⌋2−3x+2=k2−3x+2=0, which means x=3k2+2 with k<x<k+1. Because 3k2+2>0, only nonnegative values of k need to be considered. The condition k<3k2+2 is equivalent to 0<k2−3k+2, which is true if and only if k<1 or k>2. The only nonnegative value with k<1 is k=0, leading to the solution x=32. The condition 3k2+2<k+1 is equivalent to k2−3k−1<0. The roots of k2−3k−1 are k=23±13, so k2−3k−1<0 if and only if
−0.3≈23−13<k<23+13≈3.3.
The only new integer value is k=3, leading to the fourth solution x=311.
OR
Because ⌊x⌋2 is a nonnegative integer, 3x−2 must also be a nonnegative integer, so x=3n for some integer n≥2. Furthermore, if ⌊x⌋≥4, then\ ⌊x⌋2−3x+2=⌊x⌋2−3(x−1)−1≥⌊x⌋2−3⌊x⌋−1=⌊x⌋(⌊x⌋−3)−1≥⌊x⌋−1≥3>0,\
so 2≤n≤11 and ⌊x⌋2=3x−2=n−2. The only perfect square values of n−2 in this range are 0,1,4, and 9 , occurring when x=32,1,2, and 311, respectively. Each of these values of x satisfies the equation, so there are (B)4 such values.