Finding solutions to $[x]^2 + 2[x]-3x=0$

Question

Related question, I was attempting the question in an alternative way I wrote $ \{ x \} = x-[x]$, then after subbing that into the equation I got the following quadratic :

$$ [x]^2 + 2[x]-3x=0$$ I rearrange this as:

$$ [x]^2 + 2 [x] = 3x$$

Adding one:

$$ ( [x] +1)^2 = 3x+1$$

Or,

$$ [x] + 1 = \pm \sqrt{3x+1}$$

Now, I use property to take in $1$ to the box function:

$$ [1+x] = \pm \sqrt{3x+1}$$

Now , I substitute $ x+1 = u$, this leads to:

$$ [u] = \pm \sqrt{3u-2}$$

The bound on $u$ is given as $ u \geq \frac23$, it is clear that the $ [u]= - \sqrt{3u-2}$ has no solutions because left side is always positive for $u \geq \frac23$ , hence can't be equal to negative number of RHS. This leads to a single inequality:

$$ [u] = \sqrt{3u-2}$$

This leads for checking when $3u-2$ is perfect square:

$$ 3u-2 = 0$$

$$ 3u-2=1 \implies u=1$$ $$ 3u -2 = 4 \implies u=2$$ $$ 3u-2 =9 \implies u= \frac{11}{2}$$

Seems like there are no solutions for $ u >2$ , so how do I proof that this is the case?

Answer 1

$[u] = \sqrt{3u - 2}$ has no solution if $\sqrt{3u-2} < u-1$ since $[u] > u-1$.

This is equivalent to $3u-2 < u^2-2u+1$, or $u^2-5u+3 = (u-2.5)^2 - 3.25 > 0$

Hence there is no solution for $u$ if $u > \sqrt{3.25} + 2.5 \approx 4.30278$.

Answer 2

Using $\{x\}$ is advantageous. The original equation transforms to $$[x]^2-[x]=3\{ x\}$$

LHS is an integer $\Rightarrow$ $\{x\}=0,1/3,2/3$. So $x$ is $n$ or $n+1/3$ or $n+2/3$ for some integer $n$.

Substituting these in original equation gives $$n^2-n=0,1,2$$

which can be easily solved. Now there is no need to check anything. Only correct solutions will be obtained.

Answer 3

$\left[{x}\right]^{\mathrm{2}} +\mathrm{2}\left[{x}\right]−\mathrm{3}{x}=\mathrm{0} \\ $ $\left[{x}\right]^{\mathrm{2}} −\left[{x}\right]−\mathrm{3}\left\{{x}\right\}=\mathrm{0} \\ $ $\left[{x}\right]=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{12}\left\{{x}\right\}}}{\mathrm{2}}\:\in\left\{-\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}\right\} \\ $ ${x}=\frac{\left(\left[{x}\right]+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}\in\left\{\frac{-\mathrm{1}}{\mathrm{3}},\mathrm{0},\mathrm{1},\frac{\mathrm{8}}{\mathrm{3}}\right\} \\ $