Let m and n be the numbers of real roots of t | vedclass

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(A) For the first equation: $x^2-12x+[x]+31=0$. Since $x = [x] + \{x\}$,we have $x^2-12([x]+\{x\})+[x]+31=0$. Rearranging gives $\{x\} = \frac{x^2-11[

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$9$

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(A) For the first equation: $x^2-12x+[x]+31=0$. Since $x = [x] + \{x\}$,we have $x^2-12([x]+\{x\})+[x]+31=0$. Rearranging gives $\{x\} = \frac{x^2-11[x]+31}{12}$. Alternatively,$x^2-12x+31 = -[x]$. Let $[x] = k$,then $x^2-12x+31+k=0$. Solving for $x$,$x = \frac{12 \pm \sqrt{144-4(31+k)}}{2} = 6 \pm \sqrt{36-31-k} = 6 \pm \sqrt{5-k}$. For $x$ to be real,$5-k \geq 0$,so $k \leq 5$. Also,$k \leq 6 \pm \sqrt{5-k} Case $k=5$: $x = 6 \pm 0 = 6$. But $[6]=6 
eq 5$. Case $k=4$: $x = 6 \pm 1 = 7, 5$. $[7]=7 
eq 4$,$[5]=5 
eq 4$. Case $k=3$: $x = 6 \pm \sqrt{2} \approx 7.41, 4.58$. $[4.58]=4 
eq 3$. Case $k=2$: $x = 6 \pm \sqrt{3} \approx 7.73, 4.26$. $[4.26]=4 
eq 2$. Case $k=1$: $x = 6 \pm 2 = 8, 4$. $[4]=4 
eq 1$. Case $k=0$: $x = 6 \pm \sqrt{5} \approx 8.23, 3.76$. $[3.76]=3 
eq 0$. Checking $x^2-12x+[x]+31=0$ again: if $x \in [k, k+1)$,then $x^2-12x+k+31=0$. For $k=5$,$x^2-12x+36=0 \implies x=6$,not in $[5,6)$. For $k=4$,$x^2-12x+35=0 \implies (x-5)(x-7)=0 \implies x=5$,which is in $[4,5)$. So $x=5$ is a root. For $k=3$,$x^2-12x+34=0 \implies x = 6 \pm \sqrt{2}$,neither in $[3,4)$. Thus,$m=1$. For the second equation: $x^2-5|x+2|-4=0$. If $x \geq -2$,$x^2-5(x+2)-4=0 \implies x^2-5x-14=0 \implies (x-7)(x+2)=0 \implies x=7, -2$. If $x The set of roots is $\{7, -2, -3\}$,so $n=3$. Then $m^2+mn+n^2 = 1^2 + (1)(3) + 3^2 = 1+3+9 = 13$. Wait,checking the options,if $m=0, n=3$,$m^2+mn+n^2=9$. Let's re-verify $m$. $x^2-12x+[x]+31=0$. If $x=5$,$25-60+5+31=1 eq 0$. Actually,there are no real roots for the first equation,$m=0$. Then $m^2+mn+n^2 = 0^2+0(3)+3^2 = 9$.

Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^2-12x+[x]+31=0$ and $x^2-5|x+2|-4=0$ respectively,where $[x]$ denotes the greatest integer $\leq x$. Then $m^2+mn+n^2$ is equal to $..............$.

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