Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

Question

Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :- $$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$ Now, the LHS is an integer, so $2|(4x-1)$ . I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.

This is where I am stuck, how should I proceed next?

Can anyone help me?

Answer 1

As $2x-\dfrac{1}{2}$ is an integer, $x$ is either $n+\dfrac{1}{4}$ or $n+\dfrac{3}{4}$, $n \in \mathbb{Z}$. (Recall any real $x$ can be written as sum of an integer and a positive fraction).

Substitute this in original equation to solve for $n$.

$$n=-1 \Rightarrow x=-1+\dfrac{1}{4}$$
$$n=-2 \Rightarrow x=-2+\dfrac{3}{4}$$

Answer 2

Let us find a necessary condition.

$$\lfloor 3x+1\rfloor=2x-\frac 12 \implies$$

$$\lfloor 3x\rfloor=2x-\frac 32 \implies $$

$$3x-\lfloor 3x\rfloor =x+\frac 32\implies$$

$$0\le x+\frac 32<1 \implies$$

$$-\frac 32\le x<-\frac 12\implies $$ $$-\frac 72\le 2x-\frac 12<-\frac 32\implies$$

$$2x-\frac 12\in \{-3,-2\}\implies$$ $$x\in\{-\frac 54,-\frac 34\}$$

We check this condition is sufficient. the sum of the roots is $$S=-\frac 54-\frac 34=-2$$

Answer 3

$y=3x+1 \implies [y] = (4y-7)/6$.

Let $[y]=n, \{y\}=r$ then $n=(4n+4r-7)/6, 2n=4r-7$.

Since $0\le r < 1$, $0 \le 4r < 4$, we must have $4r=1,3$, so either $r=\frac 14, n=-3$ or $r=\frac 34, n=-2$.

Finally, $3x+1 = -\frac{11}{4}$ or $-\frac 54$, $x=-\frac 54$ or $-\frac 34$.

(You don't need $y$ but it simplifies things a bit.)

Answer 4

I'm leaving this here, but it's wrong. I added a correct answer below.

\begin{align} \lfloor 3x + 1 \rfloor &= 2x - \frac{1}{2} \\ \lfloor 3x \rfloor + 1 &= 2x - \frac{1}{2} \\ \lfloor 3x \rfloor &= 2x - \frac 32 \end{align}

So, for some integer, $n$,

\begin{array}{ccccc} &&\lfloor 3x \rfloor = n \\ \hline n &\le &3x &< &n+1 \\ \frac n3 &\le &x &< &\frac n3 + \frac 13 \\ \hline \frac {2n}3 &\le &2x &< &\frac {2n}3 + \frac 23 \\ \frac {2n}3 - \frac 32 &\le &2x - \frac 32 &< &\frac {2n}3 - \frac 56 \\ \frac {2n}3 - \frac 32 &\le &n &< &\frac {2n}3 - \frac 56 \\ -\frac 32 &\le &\frac n3 &< &-\frac 56 \\ -\frac 92 &\le & n &< &-\frac 52 \\ -4 &\le & n &\le &-3 \\ \hline -4 &\le & \lfloor 3x \rfloor &\le &-3 \\ -4 &\le & 2x - \frac 12 &\le &-3 \end{array}

Hence $x \in \left\{ -\dfrac 74, -\dfrac 54 \right\}$

The sum of the roots is $-3$.

This is a correct answer.

Let $x = n + \epsilon$ where $n \in \mathbb R$ and $\epsilon \in [0,1)$.

\begin{align} \lfloor 3x \rfloor &= 2x - \frac 32 \\ 3n + \lfloor 3\epsilon \rfloor &= 2n + 2\epsilon - \frac 32 \\ 6n + 2\lfloor 3\epsilon \rfloor &= 4n + 4\epsilon - 3 \\ 2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\ \end{align}

Since the LHS is an integer, then so too must $4\epsilon$ be an integer. It follows that $4\epsilon \in \{0,1,2,3\}$

If $\epsilon = 0$.

\begin{align} 2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\ 2n + 3 &= 0 \\ &\text{No solution.} \end{align}

If $\epsilon = \dfrac 14$.

\begin{align} 2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\ 2n + 3 &= 1 \\ n &= -1 \\ x &= -\dfrac 34 \end{align}

If $\epsilon = \dfrac 12$.

\begin{align} 2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\ 2n + 5 &= 2 \\ &\text{No solution.} \end{align}

If $\epsilon = \dfrac 34$.

\begin{align} 2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\ 2n + 7 &= 3 \\ n &= -2 \\ x &= -\dfrac 54 \end{align}