# If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$.

If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$ where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.

Not sure if all these denotions are conventions that always mean what I've mentioned. Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.

• You may assume that $k\le x< k+1$ with $k\in \mathbb{Z}$ and then try to solve $4x^2-40k+51=0$ over $\mathbb{R}$ and see if the solutions belong to $[k,k+1)$. – user37238 Jan 20 '14 at 15:44
• Saying "integer part of the number", as you do, is fine and standard. – ShreevatsaR Jan 20 '14 at 15:49
• And by the way, someone has added the tag 'calculus' to my question. I haven't been taught about calculus yet, so please don't use it in the solution. Can I remove the tag? – user26486 Jan 20 '14 at 15:54
• @user37238 Thanks, I've managed to solve it and got the solutions $k=2$, $k=6$, $k=7$, $k=8$. – user26486 Jan 20 '14 at 16:03
• @mathh, please create another thread for the other Question? – lab bhattacharjee Jan 20 '14 at 16:19

Since $$x-1\lt \lfloor x\rfloor \le x,$$ we have \begin{align}x-1\lt \frac{4x^2+51}{40}\le x&\iff 4x^2-40x+91\gt0\ \text{and}\ 4x^2-40x+51\le 0\\&\iff 1.5\le x\lt 3.5\ \text{or}\ 6.5\lt x\le 8.5.\end{align}

1) When $1.5\le x\lt 2\Rightarrow \lfloor x\rfloor=1$, $$4x^2-40\times 1+51=0$$ does not have any real solution.

2) When $2\le x\lt 3\Rightarrow \lfloor x\rfloor=2$, $$4x^2-40\times 2+51=0\Rightarrow x=\pm \sqrt{29}/2\Rightarrow x=\sqrt{29}/2.$$

3) When $3\le x\lt 3.5\Rightarrow \lfloor x\rfloor=3$, $$4x^2-40\times 3+51=0\Rightarrow x=\pm \sqrt{69}/2.$$ But these don't satisfy $3\le x\lt 3.5.$

4) When $6.5\lt x\lt 7\Rightarrow \lfloor x\rfloor=6$, $$4x^2-40\times 6+51=0\Rightarrow x=\pm 3\sqrt{21}/2\Rightarrow x=3\sqrt{21}/2.$$

5) When $7\le x\lt 8\Rightarrow \lfloor x\rfloor=7$, $$4x^2-40\times 7+51=0\Rightarrow x=\pm \sqrt{229}/2\Rightarrow x=\sqrt{229}/2.$$

6) When $8\le x\le 8.5\Rightarrow \lfloor x\rfloor=8$, $$4x^2-40\times 8+51=0\Rightarrow x=\pm \sqrt{269}/2\Rightarrow x=\sqrt{269}/2.$$

Hence, the answer is $$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$

So I've decided to answer my own question. Let $k\le x < k+1$ with $k\in\mathbb Z$. Then $$4x^2-40k+51=0$$

Now we need to find the solution $x$ in terms of $k$. That solution has to belong to the interval $[k\:;k+1)$. We can easily find out that $x$ is actually $$x=\pm\sqrt{10k-12.75}$$

As I said, the solution has to be in the interval, thus $$\pm\sqrt{10k-12.75}\ge k\tag{1}$$

We're only talking about real numbers, hence $$10k-12.75\ge0$$

Which shows that $$k\ge 1.275\tag{2}$$ $$\implies x>0\implies x=\sqrt{10k-12.75}$$

(I've thus removed the $\pm$ sign from the $x$)

Thus we'll only talk about positive numbers ($x,k$ are positive). This lets us square both sides of the inequality $(1)$ (remember we don't need the $\pm$ anymore): $$10k-12.75\ge k^2$$

Which shows that $k\in [1.5\:;8.5]\tag{3}$

We also know that $$\sqrt{10k-12.75}< k+1$$

And by doing the same we see that $k\in (-\infty\:;2.5)\cup(5.5\:;+\infty)\tag{4}$

$(2)(3)(4)\Rightarrow k=2, 6, 7, 8$.

We can substitute this to $x=\sqrt{10k-12.75}$ and get that $$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$

Note that when $i$ is odd then $3|(2^{i}-2)$ and hence $\lfloor \frac {2^{i}}{3} \rfloor = \lfloor \frac {2^{i}-2+2}{3} \rfloor = \lfloor \frac {2^{i}-2}{3} + \frac {2}{3} \rfloor = \frac {2^{i}-2}{3}$.
Also when $j$ is even, then $3|(2^{j}-1)$ and hence $\lfloor \frac {2^{j}}{3} \rfloor = \lfloor \frac {2^{j}-1+1}{3} \rfloor = \lfloor \frac {2^{j}-1}{3} + \frac {1}{3} \rfloor = \frac {2^{j}-1}{3}$. Therefore, $$\sum_{r=0}^{2010} \lfloor \frac {2^{r}}{3} \rfloor = \sum_i \lfloor \frac {2^{i}}{3} \rfloor + \sum_j \lfloor \frac {2^{j}}{3} \rfloor = \sum_i \frac {2^{i}-2}{3} + \sum_j \frac {2^{j}-1}{3}$$

• But now we have to evaluate $$\sum_{i=0}^{2010}2^i$$ How can we do this? – user26486 Jan 20 '14 at 16:55
• To others who hadn't seen the second problem before I deleted it: Find the value of the sum $$\sum_{i=0}^{2010}\lfloor\frac{2^i}{3}\rfloor$$ – user26486 Jan 20 '14 at 17:02
• @mathh , here the summation $i$ is over all odd numbers $\lt 2010$ and the summation $j$ is over all even numbers $\leq 2010$. The sequence $\{ a, ar, ar^{2}, ar^{3} ... \}$ is a geometric progression and the sum of first n terms is given by $\frac {a(r^{n}-1)}{r-1}$ for $r \neq 1$ ... try to prove it, its simple. – Indrayudh Roy Jan 20 '14 at 17:03
• So is the final answer $\frac{2^{2011}-1}{3}-2011$ ? – user26486 Jan 20 '14 at 17:10
• Do you mean that $$a+ar+ar^2+\ldots+ar^n=\frac{a(r^{n+1}-1)}{r-1} ?$$ And sorry but I haven't been taught about progressions yet. – user26486 Jan 20 '14 at 17:14