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  4. The solution set of (x)2+(x+1)2=25, where (x) is the least integer greater than or equal to x, is

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Q. The solution set of $(x)^{2}+(x+1)^{2}=25$, where $(x)$ is the least integer greater than or equal to $x$, is

Complex Numbers and Quadratic Equations

Solution:

If $x=n \in Z,(x)^{2}+(x+1)^{2}=25$
$\Rightarrow n^{2}+(n+1)^{2}=25$
$\Rightarrow 2 n^{2}+2 n-24=0$
$\Rightarrow n^{2}+n-12=0$
$\Rightarrow n=3,-4$
$\therefore x=3,-4$
If $x=n+k, n \in Z, 0< k < 1$, then
$(x)^{2}+(x+1)^{2}=25 $
$\Rightarrow (n+1)^{2}+(n+2)^{2}=25$
$\Rightarrow 2 n^{2}+6 n-20=0$
$\Rightarrow n^{2}+3 n-10=0 $
$\Rightarrow n=2,-5$
$\therefore x=2+k,-5+k$,
where $0< k < 1$
$\therefore x >2, x >-5$
$\therefore $ Solution set is $(-5,-4] \cup(2,3]$