Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[x]$, returns the largest integer less than or equal to $x$. We often use the property $x = [x] + \{x\}$, where $\{x\}$ is the fractional part of $x$ and $0 \le \{x\} < 1$.
• Absolute Value Function: The absolute value function, denoted by $|x|$, is defined as $x$ if $x \ge 0$ and $-x$ if $x < 0$. When solving equations involving absolute values, we consider different cases based on the sign of the expression inside the absolute value.
• Quadratic Equations: A quadratic equation of the form $ax^2 + bx + c = 0$ has real roots if its discriminant, $D = b^2 - 4ac$, is greater than or equal to zero. The quadratic $ax^2+bx+c$ is always positive if $a>0$ and $D<0$.

Step-by-Step Solution

Step 1: Solve the first equation: $x^2 - 12x + [x] + 31 = 0$

We will use the property $[x] = x - \{x\}$ to rewrite the equation in terms of the fractional part of $x$. This is done to utilize the known bounds of the fractional part, which lie between 0 and 1. $x^2 - 12x + (x - \{x\}) + 31 = 0$ $x^2 - 11x + 31 - \{x\} = 0$ $\{x\} = x^2 - 11x + 31$

Step 2: Apply the range of the fractional part: $0 \le \{x\} < 1$

Since $0 \le \{x\} < 1$, we have $0 \le x^2 - 11x + 31 < 1$. This inequality restricts the possible values of $x$.

Step 3: Solve the compound inequality: $0 \le x^2 - 11x + 31 < 1$

We split the compound inequality into two inequalities:

1. $x^2 - 11x + 31 \ge 0$
2. $x^2 - 11x + 31 < 1$

Let's solve each one:

• $x^2 - 11x + 31 \ge 0$: The discriminant is $D = (-11)^2 - 4(1)(31) = 121 - 124 = -3 < 0$. Since the leading coefficient is positive and the discriminant is negative, $x^2 - 11x + 31 > 0$ for all real $x$. Thus, this inequality is always true.

• $x^2 - 11x + 31 < 1$: $x^2 - 11x + 30 < 0$ $(x - 5)(x - 6) < 0$ The roots are $x = 5$ and $x = 6$. The inequality holds when $5 < x < 6$.

Combining the solutions, we have $5 < x < 6$.

Step 4: Determine the value of $[x]$ for $5 < x < 6$

If $5 < x < 6$, then $[x] = 5$.

Step 5: Substitute $[x] = 5$ into the original equation and solve for $x$

Substituting $[x] = 5$ into $x^2 - 12x + [x] + 31 = 0$, we get: $x^2 - 12x + 5 + 31 = 0$ $x^2 - 12x + 36 = 0$ $(x - 6)^2 = 0$ $x = 6$

Step 6: Verify the solution against the domain restriction

We found $x = 6$, but the condition from Step 3 is $5 < x < 6$. Since $x = 6$ does not satisfy $5 < x < 6$, it's not a valid solution. Thus, the first equation has no real roots, and $m = 0$.

Step 7: Solve the second equation: $x^2 - 5|x+2| - 4 = 0$

We consider two cases based on the absolute value.

• Case 1: $x + 2 \ge 0 \Rightarrow x \ge -2$ In this case, $|x+2| = x+2$. Substituting into the equation: $x^2 - 5(x+2) - 4 = 0$ $x^2 - 5x - 10 - 4 = 0$ $x^2 - 5x - 14 = 0$ $(x - 7)(x + 2) = 0$ $x = 7 \text{ or } x = -2$ Since $x \ge -2$, both $x = 7$ and $x = -2$ are valid solutions.

• Case 2: $x + 2 < 0 \Rightarrow x < -2$ In this case, $|x+2| = -(x+2)$. Substituting into the equation: $x^2 - 5(-(x+2)) - 4 = 0$ $x^2 + 5(x+2) - 4 = 0$ $x^2 + 5x + 10 - 4 = 0$ $x^2 + 5x + 6 = 0$ $(x + 2)(x + 3) = 0$ $x = -2 \text{ or } x = -3$ Since $x < -2$, only $x = -3$ is a valid solution. $x=-2$ is extraneous.

Step 8: Consolidate the roots for the second equation

The real roots for the second equation are $x = -3, -2, 7$. Therefore, $n = 3$.

Step 9: Calculate $m^2 + mn + n^2$

We have $m = 0$ and $n = 3$. $m^2 + mn + n^2 = (0)^2 + (0)(3) + (3)^2 = 0 + 0 + 9 = 9$

Common Mistakes & Tips

• Forgetting to check the domain: Always verify that the solutions obtained after solving equations with greatest integer or absolute value functions satisfy the initial domain restrictions.
• Incorrectly handling the absolute value: Remember to consider both positive and negative cases when dealing with absolute values.
• Misunderstanding the greatest integer function: Remember that $[x]$ is an integer, and use the property $x = [x] + \{x\}$ to relate $x$ to its integer and fractional parts.

Summary

The number of real roots for the first equation is $m = 0$, and the number of real roots for the second equation is $n = 3$. Therefore, $m^2 + mn + n^2 = 9$.

Final Answer

The final answer is \boxed{9}.