Solve the following for x, where |x| is the modulus function, [x] is the greatest integer function, \x\ is the

MISCELLANEOUS EXERCISE 6 (II) · Q190

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$[x^2]-5[x]+6=0 \implies [x^2]=5[x]-6$ . Let $n=[x]$ , so $n\le x<n+1$ .

Since $[x^2]\ge0$ always (as $x^2\ge0$ ), we need $5n-6\ge0 \implies n\ge1.2 \implies n\ge2$ (integer).

For $n\ge2$ , $x\ge n\ge2>0$ , so $x^2$ increases with $x$ on $[n,n+1)$ , giving $x^2\in[n^2,(n+1)^2)$ .

$n=2$ : need $[x^2]=5(2)-6=4$ , i.e. $4\le x^2<5$ . Combined with $x\in[2,3)$ : $x\in[2,\sqrt5)$ (since $\sqrt5\approx2.236<3$ ).

$n=3$ : need $[x^2]=5(3)-6=9$ , i.e. $9\le x^2<10$ . Combined with $x\in[3,4)$ : $x\in[3,\sqrt{10})$ (since $\sqrt{10}\approx3.162<4$ ). …

Q.Solve the following for $x$ , where $|x|$ is the modulus function, $[x]$ is the greatest integer function, $\{x\}$ is the fractional part function: $[x^2]-5[x]+6=0$ .