Solving Quadratic Inequalities: Find The Solution Interval

Understanding quadratic inequalities and how to solve them is a fundamental skill in algebra. This article will guide you through the process of finding the interval where the quadratic inequality $x^2 - 4x - 12 ≤ 0$ holds true. We'll break down the steps, explain the concepts, and ensure you grasp the methodology to tackle similar problems confidently.

Understanding Quadratic Inequalities

Quadratic inequalities are mathematical expressions that compare a quadratic expression to another value, typically zero. The general form of a quadratic inequality is $ax^2 + bx + c < 0$, $ax^2 + bx + c > 0$, $ax^2 + bx + c ≤ 0$, or $ax^2 + bx + c ≥ 0$, where a, b, and c are constants, and x is a variable. Solving these inequalities involves finding the range of x values that satisfy the given condition.

The inequality $x^2 - 4x - 12 ≤ 0$ is a specific instance of a quadratic inequality. Our goal is to determine the interval(s) of x values for which this inequality is true. This process typically involves factoring the quadratic expression, finding the critical points (zeros), and testing intervals to identify the solution set.

Why are quadratic inequalities important? They arise in various real-world applications, such as optimization problems, physics (e.g., projectile motion), and economics (e.g., cost analysis). Mastering the techniques to solve these inequalities provides a robust foundation for more advanced mathematical concepts and practical problem-solving.

To effectively solve quadratic inequalities, it is important to understand the behavior of quadratic functions, particularly their graphs (parabolas) and how they intersect the x-axis. The solutions to the corresponding quadratic equation (where the inequality is replaced with an equality) play a crucial role in determining the intervals where the inequality holds true. Additionally, it is essential to grasp the concept of interval notation, which is used to express the solution sets concisely. With these foundational concepts in mind, one can approach solving quadratic inequalities systematically and accurately.

Step-by-Step Solution for $x^2 - 4x - 12 ≤ 0$

To solve the quadratic inequality $x^2 - 4x - 12 ≤ 0$, we will follow a step-by-step approach that includes factoring, finding critical points, and testing intervals. This structured method ensures accuracy and clarity in the solution process.

1. Factor the Quadratic Expression

The first step is to factor the quadratic expression $x^2 - 4x - 12$. We are looking for two numbers that multiply to -12 and add to -4. These numbers are -6 and 2. Therefore, the factored form of the expression is:

$(x - 6)(x + 2) ≤ 0$

Factoring the quadratic expression simplifies the problem by allowing us to identify the critical points where the expression changes its sign. This step is fundamental because the sign of the quadratic expression is determined by the signs of its factors. When both factors have the same sign (both positive or both negative), the product is positive. When the factors have opposite signs, the product is negative. This understanding is crucial for identifying the intervals where the inequality holds true. Moreover, factoring is a widely applicable technique in algebra, and proficiency in factoring quadratic expressions is beneficial for solving various types of equations and inequalities.

2. Find the Critical Points

The critical points are the values of x that make the quadratic expression equal to zero. These points are the roots of the quadratic equation $(x - 6)(x + 2) = 0$. Setting each factor equal to zero, we get:

$x - 6 = 0$ => $x = 6$

$x + 2 = 0$ => $x = -2$

So, the critical points are $x = -2$ and $x = 6$. These critical points divide the number line into three intervals: $(-\infty, -2)$, $(-2, 6)$, and $(6, \infty)$.

The critical points are essential because they are the boundaries of the intervals where the quadratic expression can change its sign. The quadratic expression can only change its sign at these points, which are the roots of the quadratic equation. This is because a continuous function (like a quadratic) can only go from positive to negative (or vice versa) by passing through zero. Therefore, by identifying these critical points, we can systematically test each interval to determine where the quadratic expression satisfies the given inequality. This step is a crucial bridge between the algebraic manipulation (factoring) and the analysis of the intervals, allowing us to pinpoint the exact solution set.

3. Test Intervals

Now, we need to test each of the three intervals to see where $(x - 6)(x + 2) ≤ 0$.

• Interval 1: $(-\infty, -2)$

  Choose a test point, say $x = -3$. Plug it into the factored inequality:

  $(-3 - 6)(-3 + 2) = (-9)(-1) = 9 > 0$. So, the inequality does not hold true in this interval.

• Interval 2: $(-2, 6)$

  Choose a test point, say $x = 0$. Plug it into the factored inequality:

  $(0 - 6)(0 + 2) = (-6)(2) = -12 ≤ 0$. So, the inequality holds true in this interval.

• Interval 3: $(6, \infty)$

  Choose a test point, say $x = 7$. Plug it into the factored inequality:

  $(7 - 6)(7 + 2) = (1)(9) = 9 > 0$. So, the inequality does not hold true in this interval.

Testing intervals is a pivotal step in solving inequalities because it allows us to determine the sign of the expression within each interval. By selecting a representative point within each interval and substituting it into the inequality, we can deduce whether the expression is positive, negative, or zero in that interval. This method is based on the principle that the sign of a continuous function (like a quadratic) can only change at its roots (critical points). Therefore, if the inequality holds true for a test point within an interval, it holds true for the entire interval. Conversely, if the inequality does not hold true for a test point, it does not hold true for the entire interval. This systematic approach ensures that we consider all possible solutions and accurately identify the solution set.

4. Include the Critical Points

Since the inequality is $≤ 0$, we include the critical points where the expression equals zero. Thus, $x = -2$ and $x = 6$ are part of the solution.

Including the critical points in the solution set is a critical step because the inequality includes the