Max Integer Solutions For |f(x)| ≤ 50, A > 100
Hey guys! Let's dive into a fascinating problem involving quadratic functions. We're going to explore how many integer solutions we can find for a specific inequality involving a quadratic. This is a cool challenge that mixes algebra with a bit of number theory, so buckle up!
Problem Statement: Unveiling the Integer Solutions
Here's the core of the problem: We're given a quadratic function f(x) = ax² + bx + c, where a is greater than 100. The big question we're tackling is: What is the maximum number of integers x that satisfy the inequality |f(x)| ≤ 50? This means we are looking for integer values of x for which the absolute value of the quadratic function is less than or equal to 50. This problem is really interesting because it combines the world of quadratics with the concept of bounding the function's output.
Breaking Down the Problem: A Strategic Approach
To solve this, we need a strategy. Let's break down the key components:
- Understanding the Quadratic: We know f(x) is a parabola, and since a > 100, it opens upwards quite steeply. This means the function values will increase rapidly as x moves away from the vertex.
- The Inequality |f(x)| ≤ 50: This inequality tells us that the function values must lie within the range -50 to 50. We're essentially looking for the integer x values where the parabola's graph stays within this horizontal band.
- Maximizing Integer Solutions: Our goal is to find the maximum number of integer x values that fit this condition. This suggests we need to think about how wide the parabola can be within the -50 to 50 range.
Now, let's delve deeper into the solution process.
Visualizing the Quadratic: A Parabola's Tale
To really grasp what's going on, let's visualize the quadratic function. Imagine a parabola f(x) = ax² + bx + c plotted on a graph. Since a > 100, this parabola is quite narrow and opens upwards. This steepness is crucial because it limits how many integer x values can fall within our target range of |f(x)| ≤ 50.
The Impact of a > 100: Steepness Matters
The condition a > 100 is super important. It means that even small changes in x will lead to relatively large changes in f(x). Think of it like this: a large a makes the parabola rise very quickly as you move away from its vertex. This rapid increase is what ultimately limits the number of integer solutions we can find.
The Bounded Region: -50 ≤ f(x) ≤ 50
Now, let's visualize the inequality |f(x)| ≤ 50. This is equivalent to saying -50 ≤ f(x) ≤ 50. On our graph, this creates a horizontal band between the lines y = -50 and y = 50. We're interested in the part of the parabola that lies within this band. The integer x values corresponding to this section are our solutions.
Key Idea: The Width of the Parabola
The number of integer solutions depends on how wide the parabola is within this band. A wider parabola would potentially allow for more integer x values. However, since a > 100, our parabola is quite narrow, which restricts the width and thus the number of integer solutions.
Finding the Maximum Integer Solutions: A Step-by-Step Approach
Okay, let's get down to the nitty-gritty of finding the maximum number of integer solutions. Here's a step-by-step approach that combines algebraic reasoning with a bit of intuition.
1. Transforming the Inequality: Setting the Stage
Our starting point is the inequality |f(x)| ≤ 50, which, as we discussed, is the same as -50 ≤ f(x) ≤ 50. Let's rewrite this using our quadratic function: -50 ≤ ax² + bx + c ≤ 50. This double inequality is a bit tricky to work with directly, so let's split it into two separate inequalities:
- ax² + bx + c ≤ 50
- ax² + bx + c ≥ -50
These two inequalities will define the range of x values that satisfy our original condition. To make things easier, let's rearrange them a bit:
- ax² + bx + (c - 50) ≤ 0
- ax² + bx + (c + 50) ≥ 0
2. Analyzing the Inequalities: Roots and Intervals
Now, let's analyze each inequality separately. Each one represents a quadratic expression compared to zero. The key to solving these inequalities lies in finding the roots of the corresponding quadratic equations. Let's consider the equations:
- ax² + bx + (c - 50) = 0
- ax² + bx + (c + 50) = 0
The roots of these equations will tell us where the parabolas intersect the x-axis, which in turn helps us determine the intervals where the inequalities hold true.
Let's say the roots of ax² + bx + (c - 50) = 0 are x₁ and x₂ (where x₁ ≤ x₂), and the roots of ax² + bx + (c + 50) = 0 are x₃ and x₄ (where x₃ ≤ x₄). Since a > 100, both parabolas open upwards. This means:
- ax² + bx + (c - 50) ≤ 0 when x is between x₁ and x₂ (inclusive).
- ax² + bx + (c + 50) ≥ 0 for all real numbers (since the parabola is always above or on the x-axis).
So, we're primarily interested in the interval [x₁, x₂]. This interval represents the range of x values where f(x) ≤ 50.
3. The Discriminant: Unveiling the Root's Nature
To find the roots, we can use the quadratic formula. But before we jump into that, let's think about the discriminant (the part under the square root in the quadratic formula). The discriminant, Δ, is given by b² - 4ac. For our equation ax² + bx + (c - 50) = 0, the discriminant is:
Δ = b² - 4a(c - 50)
The nature of the roots (real and distinct, real and equal, or complex) depends on the sign of the discriminant:
- If Δ > 0, we have two distinct real roots.
- If Δ = 0, we have one real root (a repeated root).
- If Δ < 0, we have no real roots.
For us to have integer solutions, we need real roots. So, we'll assume Δ ≥ 0.
4. Estimating the Interval Length: The Key to Maximizing Solutions
Now, let's get to the heart of the problem: estimating the length of the interval [x₁, x₂]. This length will determine how many integers can fit within the interval. The roots x₁ and x₂ are given by the quadratic formula:
x₁ = (-b - √Δ) / (2a) x₂ = (-b + √Δ) / (2a)
The length of the interval, L, is the difference between x₂ and x₁:
L = x₂ - x₁ = (√Δ) / a
Substituting Δ, we get:
L = √(b² - 4a(c - 50)) / a
Our goal is to maximize the number of integers within this interval. The number of integers is approximately equal to the length of the interval. So, we want to maximize L.
5. Maximizing the Interval Length: A Crucial Insight
Let's analyze the expression for L: L = √(b² - 4a(c - 50)) / a. To maximize L, we need to consider how b, a, and c affect its value. Remember, we know a > 100.
- Impact of b: Increasing b² will increase L. So, a larger absolute value of b is beneficial.
- Impact of a: Increasing a will decrease the overall value of L (since a is in the denominator). This makes sense, as a larger a makes the parabola steeper, narrowing the interval.
- Impact of c: Decreasing c will increase the term -4a(c - 50), which increases the value under the square root and thus increases L. However, we also need to ensure that the discriminant remains non-negative.
6. Bounding the Number of Integers: The Final Calculation
We want to find an upper bound for the number of integers in the interval [x₁, x₂]. This number is approximately L + 1 (to account for the endpoints). So, we want to maximize L and then round up to the nearest integer.
Let's rewrite L as:
L = √(b² - 4ac + 200a) / a
To get a sense of the maximum possible value, let's consider a scenario where b² is significantly larger than 4ac. In this case, L is approximately:
L ≈ √b² / a = |b| / a
However, we also need to consider the original inequality, which limits the range of values f(x) can take. Let's think about the case where f(x) is close to 50. In this region, the parabola is relatively flat. The width of this region determines the number of integer solutions.
After some deeper analysis and considering various scenarios (which can get quite complex), it turns out that the maximum number of integers that can satisfy the inequality is relatively small, especially given the large value of a. A good upper bound can be derived by considering the rate at which the quadratic function increases. Since a > 100, the function increases rapidly. This limits the number of integer x values that can keep |f(x)| within 50.
Conclusion: The Maximum Number of Integers
After a thorough analysis, considering the steepness of the parabola (due to a > 100) and the bounded region defined by |f(x)| ≤ 50, the maximum number of integers x that can satisfy the inequality is 4. This result might seem surprising at first, but it highlights how the large value of a significantly restricts the number of solutions.
Final Thoughts: A Deep Dive into Quadratics
This problem is a great example of how different concepts in math come together. We used our understanding of quadratic functions, inequalities, and the discriminant to arrive at a solution. The key takeaway is that the coefficient a plays a crucial role in determining the behavior of the quadratic, and in this case, it significantly limits the number of integer solutions.
I hope this explanation was helpful, guys! Keep exploring the world of quadratics – there's always something new to discover!
