Problem:
Let [x] be the greatest integer less than or equal to x. Then the number of real solutions to 4x2−40⌊x⌋+51=0 is
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Let f(x)=4x2−40⌊x⌋+51 and let In​ be the interval n≤x<n+1 for integral n. Clearly f(x)>0 for x<0. For x≥0,f(x) is increasing on each interval In​ since 4x2 is increasing and −40⌊x⌋+51=−40n+51 is constant. Thus f(x) has at most one root in each In​ and such a root will exist if and only if f(n)≤0 and f(n+1−ϵ)>0 for small ϵ>0. Since f(n+1−ϵ) approaches g(n)=4(n+1)2−40n+51 as ϵ approaches 0, it suffices to check if g(n)>0 rather than checking f(n+1−ϵ) directly. Now,
So it suffices to check In​ for n=2,3,4,5,6,7,8. Checking we find g(n)>0 for n=2,6,7,8; so there are four roots.
Since 40⌊x⌋ is even, 40⌊x⌋−51 is odd, implying that 4x2 must also be an odd integer, say 2k+1, and x=2k+1​/2. Substituting in the original equation, it follows that
⌊22k+1​​⌋=20k+26​
hence one must have k=14(mod20). Furthermore,
20k+26​≤22k+1​​<20k+26​+1
Treating the two inequalities separately, multiplying by 20, squaring and completing the square, one obtains (k−74)2≤702 and (k−54)2>302. Since x2 must be positive, k is nonnegative, and it follows from the first inequality that 4≤k≤144, and from the second one that either k<24 or k>84. Putting these together, one finds that either 4≤k<24 or 84<k≤144. In these intervals the only values of k for which k=14(mod20) are k=14,94,114,134, yielding the four solutions