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

1. Find $$m$$ (Real roots of $$x^2 - 12x + [x] + 31 = 0$$)

Let $$[x] = I$$ and the fractional part be $$f$$, where $$x = I + f$$ and $$0 \le f < 1$$.

The equation becomes:

$$x^2 - 12x + 31 = -I$$

From the quadratic formula for $$x^2 - 12x + (31+I) = 0$$:

$$x = \frac{12 \pm \sqrt{144 - 4(31+I)}}{2} = 6 \pm \sqrt{36 - (31+I)} = 6 \pm \sqrt{5-I}$$

For $$x$$ to be real, $$5-I \ge 0 \implies I \le 5$$.

• If $$I=3$$, $$x = 6 \pm \sqrt{2} \approx 7.41$$ or $$4.58$$. Neither has $$[x]=3$$.
• If $$I=4$$, $$x = 6 \pm 1 = 7$$ or $$5$$. Neither has $$[x]=4$$.
• If $$I=5$$, $$x = 6 \pm 0 = 6$$. Here $$[x]=6 \neq 5$$.

Testing values shows there are no real roots for this equation. Thus, $$m = 0$$.

2. Find $$n$$ (Real roots of $$x^2 - 5|x + 2| - 4 = 0$$)

We check two cases for the absolute value:

• Case 1 ($$x \ge -2$$): $$x^2 - 5(x + 2) - 4 = 0 \implies x^2 - 5x - 14 = 0$$

Factor: $$(x-7)(x+2) = 0 \implies x = 7, -2$$. (Both valid as $$\ge -2$$)

• Case 2 ($$x < -2$$): $$x^2 + 5(x + 2) - 4 = 0 \implies x^2 + 5x + 6 = 0$$

Factor: $$(x+2)(x+3) = 0 \implies x = -2, -3$$.

$$x = -3$$ is valid ($$< -2$$). $$x = -2$$ was already found.

Distinct roots are $$\{7, -2, -3\}$$, so $$n = 3$$.

3. Final Calculation

Substitute $$m = 0$$ and $$n = 3$$ into the expression $$m^2 + mn + n^2$$:

$$0^2 + (0)(3) + 3^2 = 0 + 0 + 9 = \mathbf{9}$$

Correct Option: (A)