Solve (x^2-4)/(x^2-5x+6) < 0: Critical Points Guide

When tackling inequalities, especially those involving rational expressions like $\frac{x^2-4}{x^2-5 x+6}<0$, our primary goal is to identify the critical points. These are the values of $x$ where the expression might change its sign. For a rational function, these critical points occur at the roots of the numerator and the roots of the denominator. Understanding these points is absolutely crucial because they divide the number line into intervals, and within each interval, the sign of the expression remains constant. So, let's dive deep into finding these pivotal values and using them to solve our inequality. We'll break down the numerator and denominator separately, find their respective roots, and then assemble them to understand the behavior of the entire expression. This systematic approach ensures we don't miss any crucial detail and can confidently determine the solution set for the inequality. Think of these critical points as the signposts on the road of the number line, guiding us to the regions where our inequality holds true.

Finding the Roots of the Numerator

To begin our quest for critical points for the inequality $\frac{x^2-4}{x^2-5 x+6}<0$, we must first scrutinize the numerator: $x^2-4$. The roots of the numerator are the values of $x$ that make the numerator equal to zero. Setting $x^2-4 = 0$, we can easily solve for $x$. This is a classic difference of squares, which factors into $(x-2)(x+2)$. Therefore, the roots of the numerator are $x=2$ and $x=-2$. These are significant because, at these values, the entire fraction becomes zero. However, our inequality demands that the expression be strictly less than zero ($<0$), meaning we are looking for regions where the expression is negative, not zero. So, while $x=2$ and $x=-2$ are important numbers to consider, they will not be included in our final solution set because they make the expression equal to zero. It's a common pitfall to include roots that result in equality when the inequality is strict. Keep this distinction in mind as we move forward; these are potential boundaries where the sign could change.

Determining the Roots of the Denominator

Next, we turn our attention to the denominator of the rational expression: $x^2-5x+6$. The roots of the denominator are equally vital as critical points, but with a crucial difference in their implication. These are the values of $x$ that make the denominator equal to zero. When the denominator is zero, the expression is undefined. For the inequality $\frac{x^2-4}{x^2-5 x+6}<0$, we must exclude any value of $x$ that makes the denominator zero from our solution set, as division by zero is mathematically impermissible. To find these roots, we set $x^2-5x+6 = 0$. This is a quadratic equation that can be solved by factoring. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Thus, the denominator factors into $(x-2)(x-3)$. Setting each factor to zero, we find that the roots of the denominator are $x=2$ and $x=3$. As mentioned, these values make the expression undefined, so they will always be excluded from the solution set, regardless of whether the inequality is strict or includes equality. The root $x=2$ is particularly interesting because it's also a root of the numerator. We'll discuss the implications of this shortly, but for now, remember that $x=2$ and $x=3$ are the points where our function breaks down.

Assembling the Critical Points and Analyzing Intervals

Now, let's bring together all the critical points we've found. The critical points are the roots of the numerator ($x=-2$, $x=2$) and the roots of the denominator ($x=2$, $x=3$). We arrange these distinct values in ascending order on the number line: $-2, 2, 3$. These points divide the number line into four intervals: $(-\infty, -2)$, $(-2, 2)$, $(2, 3)$, and $(3, \infty)$. Our task is to determine the sign of the expression $\frac{x^2-4}{x^2-5 x+6}$ within each of these intervals. We can do this by picking a test value within each interval and substituting it into the expression. A key observation here is that $x=2$ is a root for both the numerator and the denominator. This means that the factor $(x-2)$ appears in both, and we can simplify the expression: $\frac{(x-2)(x+2)}{(x-2)(x-3)} = \frac{x+2}{x-3}$, provided $x \neq 2$. This simplification is very important. The original inequality $\frac{x^2-4}{x^2-5 x+6}<0$ is equivalent to $\frac{x+2}{x-3}<0$ for all $x$ except $x=2$ and $x=3$. The critical points for this simplified inequality are $x=-2$ (from the numerator) and $x=3$ (from the denominator). These are the points that define the sign changes for $\frac{x+2}{x-3}$. Let's re-evaluate using these simplified critical points: $-2$ and $3$. These divide the number line into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. We must also remember that $x=2$ is still an excluded point from the original expression's domain, even though it's not a critical point in the simplified form.

Interval 1: $(-\infty, -2)$

Let's pick a test value, say $x=-3$. Substituting into $\frac{x+2}{x-3}$: $\frac{-3+2}{-3-3} = \frac{-1}{-6} = \frac{1}{6}$. This is positive ($>0$). So, the expression is positive in this interval.

Interval 2: $(-2, 3)$

This interval needs careful handling because it contains the point $x=2$, where the original expression is undefined. We can split this interval into two: $(-2, 2)$ and $(2, 3)$.

• For the interval $(-2, 2)$: Let's pick $x=0$. Substituting into $\frac{x+2}{x-3}$: $\frac{0+2}{0-3} = \frac{2}{-3} = -\frac{2}{3}$. This is negative ($<0$). So, the expression is negative in this interval.

• For the interval $(2, 3)$: Let's pick $x=2.5$. Substituting into $\frac{x+2}{x-3}$: $\frac{2.5+2}{2.5-3} = \frac{4.5}{-0.5} = -9$. This is negative ($<0$). So, the expression is negative in this interval.

Interval 3: $(3, \infty)$

Let's pick a test value, say $x=4$. Substituting into $\frac{x+2}{x-3}$: $\frac{4+2}{4-3} = \frac{6}{1} = 6$. This is positive ($>0$). So, the expression is positive in this interval.

Conclusion: The Solution Set

We are looking for the intervals where $\frac{x^2-4}{x^2-5 x+6}<0$. Based on our analysis of the simplified expression $\frac{x+2}{x-3}<0$, we found that the expression is negative in the intervals $(-2, 2)$ and $(2, 3)$. We must also consider the constraints from the original inequality. The roots of the numerator ($x=-2, x=2$) are not included because the inequality is strict. The roots of the denominator ($x=2, x=3$) are always excluded because they make the expression undefined. Therefore, the solution set includes all $x$ values in $(-2, 2)$ and $(2, 3)$. Combining these, the solution is the union of these two intervals: $(-2, 2) \cup (2, 3)$.

This means that for any value of $x$ between $-2$ and $2$ (exclusive of $-2$ and $2$), and for any value of $x$ between $2$ and $3$ (exclusive of $2$ and $3$), the inequality $\frac{x^2-4}{x^2-5 x+6}<0$ will be true. It is essential to remember the simplification step and the implications of common roots in rational inequalities. The common root $x=2$ made the original expression indeterminate at that point, even though it simplified out. This is why $x=2$ must be excluded from the final answer. The critical points from the simplified expression ($x=-2$ and $x=3$) correctly guide the sign changes, but the domain restrictions from the original denominator ($x=2$ and $x=3$) and numerator ($x=2$ and $x=-2$) must always be respected.

For further exploration into solving inequalities and understanding critical points, you can refer to resources like Khan Academy for detailed lessons and practice problems on algebraic inequalities and rational functions. Paul's Online Math Notes also provides comprehensive explanations and examples that can deepen your understanding of these mathematical concepts.