Problem:
Let [x] be the greatest integer less than or equal to x. Then the number of real solutions to 4x2β40βxβ+51=0 is
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Let f(x)=4x2β40βxβ+51 and let Inβ be the interval nβ€x<n+1 for integral n. Clearly f(x)>0 for x<0. For xβ₯0,f(x) is increasing on each interval Inβ since 4x2 is increasing and β40βxβ+51=β40n+51 is constant. Thus f(x) has at most one root in each Inβ and such a root will exist if and only if f(n)β€0 and f(n+1βΟ΅)>0 for small Ο΅>0. Since f(n+1βΟ΅) approaches g(n)=4(n+1)2β40n+51 as Ο΅ approaches 0, it suffices to check if g(n)>0 rather than checking f(n+1βΟ΅) directly. Now,
So it suffices to check Inβ 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 40βxβ is even, 40βxββ51 is odd, implying that 4x2 must also be an odd integer, say 2k+1, and x=2k+1β/2. Substituting in the original equation, it follows that
β22k+1βββ=20k+26β
hence one must have k=14(mod20). Furthermore,
20k+26ββ€22k+1ββ<20k+26β+1
Treating the two inequalities separately, multiplying by 20, squaring and completing the square, one obtains (kβ74)2β€702 and (kβ54)2>302. Since x2 must be positive, k is nonnegative, and it follows from the first inequality that 4β€kβ€144, and from the second one that either k<24 or k>84. Putting these together, one finds that either 4β€k<24 or 84<kβ€144. In these intervals the only values of k for which k=14(mod20) are k=14,94,114,134, yielding the four solutions