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

Let's solve the problem step by step. 

First Equation: \( x^2 - 12x + \lfloor x \rfloor + 31 = 0 \) 
We are given the quadratic equation: \[ x^2 - 12x + \lfloor x \rfloor + 31 = 0 \] Here, \( \lfloor x \rfloor \) represents the greatest integer less than or equal to \( x \). This makes the equation non-trivial because the greatest integer function is involved, and we need to consider different intervals for \( x \) based on its integer value. 

Case 1: \( \lfloor x \rfloor = 0 \) 
If \( x \) is in the range \( [0, 1) \), then \( \lfloor x \rfloor = 0 \). The equation becomes: \[ x^2 - 12x + 31 = 0. \] Solving the quadratic equation \( x^2 - 12x + 31 = 0 \) using the quadratic formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(31)}}{2(1)} = \frac{12 \pm \sqrt{144 - 124}}{2} = \frac{12 \pm \sqrt{20}}{2}. \] Simplifying: \[ x = \frac{12 \pm 2\sqrt{5}}{2} = 6 \pm \sqrt{5}. \] We get two roots: \[ x_1 = 6 + \sqrt{5} \approx 8.236, \quad x_2 = 6 - \sqrt{5} \approx 3.764. \] Since \( x \in [0, 1) \), neither of these roots are within this interval, so there are no real roots for this case. 

Case 2: \( \lfloor x \rfloor = 1 \) 
For \( x \in [1, 2) \), \( \lfloor x \rfloor = 1 \). The equation becomes: \[ x^2 - 12x + 1 + 31 = 0 \quad \Rightarrow \quad x^2 - 12x + 32 = 0. \] Solving this quadratic equation: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2}. \] Simplifying: \[ x = \frac{12 \pm 4}{2} = 8 \text{ or } 4. \] Both roots, 8 and 4, are outside the interval \( [1, 2) \), so there are no real roots for this case either. 

Case 3: \( \lfloor x \rfloor = 2 \) 
For \( x \in [2, 3) \), \( \lfloor x \rfloor = 2 \). The equation becomes: \[ x^2 - 12x + 2 + 31 = 0 \quad \Rightarrow \quad x^2 - 12x + 33 = 0. \] Solving this quadratic equation: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(33)}}{2(1)} = \frac{12 \pm \sqrt{144 - 132}}{2} = \frac{12 \pm \sqrt{12}}{2} = \frac{12 \pm 2\sqrt{3}}{2}. \] Simplifying: \[ x = 6 \pm \sqrt{3}. \] Thus, the roots are: \[ x_1 = 6 + \sqrt{3} \approx 7.732, \quad x_2 = 6 - \sqrt{3} \approx 4.268. \] Both roots are outside the interval \( [2, 3) \), so there are no real roots for this case. 

Case 4: \( \lfloor x \rfloor = 3 \) 
For \( x \in [3, 4) \), \( \lfloor x \rfloor = 3 \). The equation becomes: \[ x^2 - 12x + 3 + 31 = 0 \quad \Rightarrow \quad x^2 - 12x + 34 = 0. \] Solving this quadratic equation: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(34)}}{2(1)} = \frac{12 \pm \sqrt{144 - 136}}{2} = \frac{12 \pm \sqrt{8}}{2} = \frac{12 \pm 2\sqrt{2}}{2}. \] Simplifying: \[ x = 6 \pm \sqrt{2}. \] Thus, the roots are: \[ x_1 = 6 + \sqrt{2} \approx 7.414, \quad x_2 = 6 - \sqrt{2} \approx 4.586. \] Both roots are outside the interval \( [3, 4) \), so there are no real roots for this case. 

Second Equation: \( x^2 - 5|x + 2| - 4 = 0 \) 
Now, we solve the second quadratic equation \( x^2 - 5|x + 2| - 4 = 0 \). 

Case 1: \( x + 2 \geq 0 \) (i.e., \( x \geq -2 \)) 
In this case, \( |x + 2| = x + 2 \). The equation becomes: \[ x^2 - 5(x + 2) - 4 = 0 \quad \Rightarrow \quad x^2 - 5x - 10 - 4 = 0 \quad \Rightarrow \quad x^2 - 5x - 14 = 0. \] Solving this quadratic equation: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2}. \] Thus, the roots are: \[ x_1 = \frac{5 + 9}{2} = 7, \quad x_2 = \frac{5 - 9}{2} = -2. \] Both roots are valid since \( x \geq -2 \). 

Case 2: \( x + 2 < 0 \) (i.e., \( x < -2 \)) 
In this case, \( |x + 2| = -(x + 2) \). The equation becomes: \[ x^2 - 5(-x - 2) - 4 = 0 \quad \Rightarrow \quad x^2 + 5x + 10 - 4 = 0 \quad \Rightarrow \quad x^2 + 5x + 6 = 0. \] Solving this quadratic equation: \[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}. \] Thus, the roots are: \[ x_1 = \frac{-5 + 1}{2} = -2, \quad x_2 = \frac{-5 - 1}{2} = -3. \] The root \( x = -2 \) is not valid because it does not satisfy \( x < -2 \), but \( x = -3 \) is valid. 

Final Calculation: \( m^2 + mn + n^2 \) 
- From the first equation, we determined that there are no real roots. - From the second equation, we have 3 roots: \( x = 7 \), \( x = -2 \), and \( x = -3 \). So \( n = 3 \). Finally, the value of \( m^2 + mn + n^2 \) is: \[ m^2 + mn + n^2 = 0 + 0 \cdot 3 + 3^2 = 9. \] Thus, the answer is: \[ \boxed{9}. \]