1
$\begingroup$

Find all solutions to $$[x^2]+2[x]=3x\text{ where } 0\le x\le 2$$ and $[x]=\lfloor x\rfloor$

$$$$ I managed to simplify this to $[x^2]-[x]=3\{x\}$. Thus, $$[x^2]-[x]=\{0,1,2\}$$ However I got stuck here and was unable to proceed further. Somehow I have a feeling that there's a much neater approach than what I've taken.$$$$ Any help would be greatly appreciated. Many thanks!

$\endgroup$
9
  • $\begingroup$ If $\left[\cdots\right]$ means floor function, the usual notation is $\left\lfloor\cdots\right\rfloor$ which is the $\LaTeX$ $\verb!\left\lfloor yourText \right\rfloor!$ $\endgroup$ Commented Jun 11, 2016 at 2:21
  • $\begingroup$ @FelixMarin I've just edited the question. $\endgroup$
    – user342209
    Commented Jun 11, 2016 at 2:29
  • $\begingroup$ In the last 24 hours you have posted quite a few questions all on the same theme: solve completely unmotivated equations involving floor functions and fractional parts of numbers. Could you please explain where in the world all these equations are coming from and why you are interested in such equations? $\endgroup$
    – KCd
    Commented Jun 11, 2016 at 3:13
  • 1
    $\begingroup$ @KCd Sir actually the last few days I had been revising Functions - I have an exam in a month which would test several chapters including Functions, and I came across a Worksheet which had about 31 questions solely on the Greatest Integer and Fractional Part Functions. Hence I thought of solving the worksheet for practice, and thus have been asking for help whenever I've been unable to solve the questions. $\endgroup$
    – user342209
    Commented Jun 11, 2016 at 3:21
  • 1
    $\begingroup$ The left side is an integer. So $3x$ is an integer. Thus the only candidates are $x=0,1/3,2/3,1,4/3,5/3,2$. Check by computing which ones work. $\endgroup$ Commented Jun 11, 2016 at 3:52

2 Answers 2

0
$\begingroup$

Write $$x={k\over3}\qquad(0\leq k\leq 6)$$ with integer $k$. The condition then is $$\left\lfloor{k^2\over 9}\right\rfloor+2\left\lfloor{k\over 3}\right\rfloor=k\ .$$ Within the given range only $k=0$ and $k=3$ do qualify, so that we obtain the two solutions $x_1=0$ and $x_2=1$.

$\endgroup$
0
$\begingroup$

If $0\leq x<1$, $LHD=0$.

$1\leq x<\sqrt2$, $LHD=3$

$\sqrt2\leq x<\sqrt3,LHD=4$

$\sqrt3\leq x<2,LHD=5$

So x is a rational number by feature of equation, $x=0,1,2$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .