Equation in real numbers // Ligning i reelle tal

 It's problem E : 17 361  in the picture.

 In translation : "E:17361.  Solve the equation  $[x]=\sqrt{\{x\}\cdot (x+1)}$  in the set of

                            real numbers.

 [Author]  Ionuţ ALEXUC Timişoara"

Remark CiP :  No matter how often they are used, the notations  $\{x\},\;[x]$  no longer have any presentation.  : {x} is fractional part and [x] is the integer part of the real number x.

In my answer S means the set of solutions to the given equation. The problem is proposed for 7th grade (students aged 12-14).

ANSWER CiP

$S=\{0,\;\sqrt{2}\}$

               

                               Solution CiP

               The domain on which the equation is defined is the interval  $x\in [-1,\;+\infty)$. (Because  $\{x\}\geqslant 0$)

          But if  $x<0$ , then  $[x]\leqslant -1$,  so strictly negative and then the equation cannot have such solutions. It remains that  $x\geqslant 0$.

           For  $0\leqslant x<1$  we have  $[x]=0$, and how  $x+1\neq 0$ it must  $\{x\}=0$, so  $x=0$.

           For  $x\geqslant 1$ , from  $x-1<[x]\leqslant x\;\Rightarrow\;\;x-1<\sqrt{\{x\}\cdot (x+1)}\leqslant x$

Hence  $(x-1)^2<\{x\}\cdot (x+1)\;\;\Rightarrow\;\{x\}>\frac{(x-1)^2}{x+1}$. But  $\{x\}<1$  so  $1>\frac{(x-1)^2}{x+1}$

which is equivalent to  $x+1>x^2-2x+1\;\Leftrightarrow\;3x>x^2\;\Leftrightarrow\;3>x$.

          So we can have  $[x]=1\;or\;[x]=2$.

          If  $[x]=1$  the equation is written  $1=\sqrt{\{x\}\cdot (1+\{x\}+1)}$

$\Leftrightarrow\;1=\{x\}\cdot(\{x\}+2)\;\Leftrightarrow\;\{x\}^2+2\{x\}=1\;\Leftrightarrow\;(\{x\}+1)^2=2\;\Leftrightarrow\;\{x\}+1=\sqrt{2}.$

Hence  $x=[x]+\{x\}=1+\{x\}=\sqrt{2}$.

          If  $[x]=2$  the equation is written  $2=\sqrt{\{x\}\cdot (2+\{x\}+1)}$

$\Leftrightarrow\;4=\{x\}\cdot (\{x\}+3)\;\Leftrightarrow\;\{x\}^2+3\{x\}=4\;\Rightarrow\;\{x\}=1$  what is not possible.

Both values ​​found for  $x$  verify the equation.

$\blacksquare$