Question

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 less than or equal to \( x \). Then \( m^2 + mn + n^2 \) is equal to ___________.

Hint:

When solving equations involving greatest integer functions or absolute values, analyze the problem by breaking it into appropriate cases based on the conditions of the functions.

Answer 1

Correct Answer: 19

1. Analysis of the first equation \( x^2 - 12x + [x] + 31 = 0 \):

- Here, \( [x] \) represents the greatest integer less than or equal to \( x \). Let \( [x] = k \), where \( k \in \mathbb{Z} \) and \( k \leq x < k+1 \).

- The equation becomes:

\[ x^2 - 12x + k + 31 = 0. \]

- For \( x \) to be a real root, the discriminant \( \Delta \) must be non-negative:

\[ \Delta = (-12)^2 - 4(1)(k+31) = 144 - 4(k+31) = 144 - 4k - 124 = 20 - 4k. \]

- For \( \Delta ≥ 0 \), we require:

\[ 20 - 4k ≥ 0 \quad \Rightarrow \quad k ≤ 5. \]

- Additionally, since \( x \) satisfies \( k ≤ x < k+1 \), and \( \Delta ≥ 0 \), we must test integer values of \( k ≤ 5 \).

Testing \( k = 5 \): The equation becomes:

\[ x^2 - 12x + 36 = 0 \quad \Rightarrow \quad (x - 6)^2 = 0. \]

This has a single root \( x = 6 \), but since \( 5 ≤ x < 6 \), no solution exists in this interval.

Hence, there are no real roots for \( x^2 - 12x + [x] + 31 = 0 \). Thus, \( m = 0 \).

2. Analysis of the second equation \( x^2 - 5|x+2| - 4 = 0 \):

- Split into cases based on \( |x+2| \):

- Case 1: \( x + 2 ≥ 0 \) (i.e., \( x ≥ -2 \)):

\[ |x+2| = x+2, \quad \text{so the equation becomes:} \]

\[ x^2 - 5(x+2) - 4 = 0 \quad \Rightarrow \quad x^2 - 5x - 10 - 4 = 0. \]

\[ x^2 - 5x - 14 = 0. \]

The discriminant is:

\[ \Delta = (-5)^2 - 4(1)(-14) = 25 + 56 = 81. \]

Roots are:

\[ x = \frac{-(-5) + \sqrt{81}}{2(1)} = \frac{5 + 9}{2}. \]

\[ x = 7 \quad \text{and} \quad x = -2. \]

Valid roots: \( x = 7 \) (as \( x ≥ -2 \)).

- Case 2: \( x + 2 < 0 \) (i.e., \( x < -2 \)):

\[ |x+2| = -(x+2), \quad \text{so the equation becomes:} \]

\[ x^2 - 5(-x-2) - 4 = 0 \quad \Rightarrow \quad x^2 + 5x + 10 - 4 = 0. \]

\[ x^2 + 5x + 6 = 0. \]

The discriminant is:

\[ \Delta = 5^2 - 4(1)(6) = 25 - 24 = 1. \]

Roots are:

\[ x = \frac{-5 + \sqrt{1}}{2(1)} = \frac{-5 + 1}{2}. \]

\[ x = -2 \quad \text{and} \quad x = -3. \]

Valid roots: \( x = -3 \) (as \( x < -2 \)).

- Combining both cases, the real roots of the second equation are:

\[ x = \{-3, -2, 7\}. \]

Thus, \( n = 3 \).

Final Answer:

\[ 9 \, . \]