Let m and n be the numbers of real roots of the quadratic equations x² - 12x + [x] + 31 = 0 and x² -

Q: Wait, why do we reject $x = 6$ in the first equation?

That is a brilliant observation! In the interval $x \in [5, 6)$, we assumed $[x] = 5$. When we solved $(x - 6)^2 - 5 = -5$, we found $x = 6$. However, our assumption was that $x$ must be in the interval $[5, 6)$, which is open at $6$. Since $6$ is not in $[5, 6)$, the intersection point does not exist. It is a classic 'hole' in the graph!

Q: Why can't I just square both sides of the modulus equation to get rid of the absolute value?

Squaring both sides of an equation involving a modulus is a dangerous game. It often introduces extraneous roots that don't satisfy the original equation. It is far safer and more rigorous to use the definition of the modulus: $|x + 2| = x + 2$ when $x \ge -2$ and $-(x + 2)$ when $x < -2$. This ensures you only find valid, real roots.

Q: How do I know which intervals to check for the modulus?

Always look for the 'critical point' where the expression inside the modulus becomes zero. Here, $x + 2 = 0$ at $x = -2$. This point divides the number line into two regions: everything to the right ($x \ge -2$) and everything to the left ($x < -2$). By checking these two regions, you cover all possible behaviors of the function.

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 $\le x$. Then $m^2 + mn + n^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey to understand the behavior of functions. The JEE Advanced exam loves to test your ability to look past the symbols and see the geometry underneath.

We have two equations, and our goal is to find the number of real roots for each. Let us peel back the layers.

### The Parabola and the Step Function

Our first equation is $x^2 - 12x + [x] + 31 = 0$. At first glance, it looks like a standard quadratic, but that $[x]$—the greatest integer function—changes everything. It is not a smooth curve; it is a staircase.

To make sense of this, let us rearrange the terms to isolate the functions:
$$x^2 - 12x + 31 = -[x]$$

Now, let us complete the square on the left side. We get:
$$(x - 6)^2 - 5 = -[x]$$

This is beautiful! The left side is a parabola with its vertex at $(6, -5)$. The right side is a step function, $-[x]$. Imagine the parabola opening upwards from $(6, -5)$.

Now, imagine the step function, which jumps at every integer value of $x$. For an intersection to occur, the parabola must hit one of these steps.

Let us test the interval $x \in [5, 6)$. In this region, $[x] = 5$, so $-[x] = -5$. Our equation becomes:
$$(x - 6)^2 - 5 = -5$$

This simplifies to $(x - 6)^2 = 0$, giving us $x = 6$. But wait! Our interval was $[5, 6)$, which is open at $6$. The value $x = 6$ is not in our interval.

This means the parabola and the step function never actually touch here; there is a 'hole' at the intersection point. If you check other intervals, you will find that the parabola grows too quickly or stays too high to ever intersect the steps. Thus, the number of real roots for the first equation is $m = 0$.

### The Modulus Chameleon

Now, let us turn our attention to the second equation: $x^2 - 5|x + 2| - 4 = 0$. The modulus function $|x + 2|$ is like a chameleon; it changes its identity based on the value of $x$. The critical point is $x = -2$.

**Case 1: $x \ge -2$**
Here, $|x + 2| = x + 2$. The equation becomes $x^2 - 5(x + 2) - 4 = 0$, which simplifies to:
$$x^2 - 5x - 14 = 0$$

Factoring this, we get $(x - 7)(x + 2) = 0$. The roots are $x = 7$ and $x = -2$. Both are $\ge -2$, so both are valid!

**Case 2: $x < -2$**
Here, $|x + 2| = -(x + 2)$. The equation becomes $x^2 - 5(-(x + 2)) - 4 = 0$, which simplifies to:
$$x^2 + 5x + 6 = 0$$

Factoring this, we get $(x + 2)(x + 3) = 0$. The roots are $x = -2$ and $x = -3$. But our condition for this case is $x < -2$. Therefore, we must reject $x = -2$ and keep only $x = -3$.

Counting our valid roots, we have $\{-3, -2, 7\}$. That gives us $n = 3$ real roots.

### The Synthesis

We have arrived at the final destination. We found $m = 0$ and $n = 3$. The problem asks us to evaluate $m^2 + mn + n^2$.

Substituting our values, we get:
$$0^2 + (0)(3) + 3^2 = 0 + 0 + 9 = 9$$

There you have it! The complexity of the problem was merely a mask for the elegance of the underlying logic. The final answer is **9**.

Animated Solution for Mathematics - Quadratic Equations: Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^{2} - 12 x + [x] + 31 = 0$ 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

Enter Numerical Value:

Visualized Solution

Problem Overview

Analyzing Equation 1

Bounding the Roots

Checking the Intersection

Conclusion for Equation 1

Analyzing Equation 2

Case 1: $x \geq - 2$

Solving Case 1

Case 2: $x < - 2$

Solving Case 2

Finding $n$

Final Calculation

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

The Parabola and the Step Function

The Modulus Chameleon

The Synthesis

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