Question

If \(\frac{x^{2}+x-12}{x^{2}-3 x+2}<0\), then \(x\) lies in (1) \((-4,3)\) (2) \((-4,2)\) (3) \([-4,1] \cup[2,3]\) (4) \((-4,1) \cup(2,3)\)

Short answer

Based on the solution, create a short answer question: Question: Find the range of values of x for which the inequality \(\frac{x^{2}+x-12}{x^{2}-3 x+2} < 0\) holds true. Answer: \((-4,1) \cup (2,3)\)

Step-by-step solution

Factorize the given expression

First, factorize the numerator and the denominator of the given fraction: Numerator: \(x^2 + x - 12\) can be factored as \((x + 4)(x - 3)\). Denominator: \(x^2 - 3x + 2\) can be factored as \((x - 1)(x - 2)\). After factorization, the expression becomes: \(\frac{(x + 4)(x - 3)}{(x - 1)(x - 2)}\).

Find the critical points of the expression

The critical points are the roots and the undefined points (where the denominator equals zero) of the expression. Roots of the expression: \((x + 4)(x - 3) = 0\), we get \(x = -4\) and \(x = 3\). Undefined points: \((x - 1)(x - 2) = 0\), we get \(x = 1\) and \(x = 2\).

Determine the sign of the expression in each interval

Arrange the critical points in ascending order: \(-4\), \(1\), \(2\), \(3\). These points divide the number line into \(5\) intervals: \((-\infty, -4)\), \((-4, 1)\), \((1, 2)\), \((2, 3)\), \((3, \infty)\). We need to test a point from each interval to check if the expression is positive or negative in that interval. Interval 1: \((-\infty, -4)\), test point: \(x = -5\). \(\frac{(-5 + 4)(-5 - 3)}{(-5 - 1)(-5 - 2)} = \frac{(-1)(-8)}{(-6)(-7)} > 0\) Interval 2: \((-4, 1)\), test point: \(x = 0\). \(\frac{(0 + 4)(0 - 3)}{(0 - 1)(0 - 2)} = \frac{(4)(-3)}{(-1)(-2)} < 0\) Interval 3: \((1, 2)\), test point: \(x = \frac{3}{2}\) \(\frac{(\frac{3}{2} + 4)(\frac{3}{2} - 3)}{(\frac{3}{2} - 1)(\frac{3}{2} - 2)} = \frac{(\frac{11}{2})(-\frac{3}{2})}{(\frac{1}{2})(-\frac{1}{2})} > 0\) Interval 4: \((2, 3)\), test point: \(x = \frac{5}{2}\) \(\frac{(\frac{5}{2} + 4)(\frac{5}{2} - 3)}{(\frac{5}{2} - 1)(\frac{5}{2} - 2)} = \frac{(\frac{13}{2})(\frac{1}{2})}{(\frac{3}{2})(\frac{1}{2})} < 0\) Interval 5: \((3, \infty)\), test point: \(x = 4\) \(\frac{(4 + 4)(4 - 3)}{(4 - 1)(4 - 2)} = \frac{(8)(1)}{(3)(2)} > 0\)

Identify the intervals where the expression is negative and select the right option

We found that the given expression is negative in intervals \((-4,1)\) and \((2,3)\). Therefore, \(x\) lies in \((-4,1) \cup (2,3)\). This matches option 4. The correct answer is: (4) \((-4,1) \cup(2,3)\)

Key concepts

Rational Expressions

Rational expressions are fractions that involve polynomials in the numerator, the denominator, or both. Like the fractions with numbers we use in everyday life, rational expressions can be simplified, added, subtracted, multiplied, and divided, as long as we're careful not to divide by zero.

Understanding rational expressions is essential when solving inequalities, as it requires us to consider where these expressions are defined and their behavior across their domain. For instance, in the given problem, the expression \(\frac{x^{2}+x-12}{x^{2}-3x+2}\) is a rational expression where the critical task is to determine for what values of \(x\) this expression is less than zero. To analyze such expressions, we utilize other mathematical concepts, including critical points, the sign test, and factorization which help in simplifying and evaluating the inequality.

Critical Points

In mathematics, critical points of a function are where its derivative is zero or undefined. However, in the context of solving inequalities with rational expressions, the term 'critical points' typically refers to the values that make the numerator zero (the roots) and the values that make the denominator zero (the undefined points).

Identifying Critical Points

The characteristics of critical points are especially important in solving inequalities because they divide the number line into distinct intervals, which can be analyzed separately. For example, in our exercise, the critical points for the expression \(\frac{x^{2}+x-12}{x^{2}-3x+2}\) are \(x=-4\), \(x=3\) from the numerator and \(x=1\), \(x=2\) from the denominator. These points lead to the boundary of intervals where the expression's sign may change, thus influencing the solution to the inequality.

Sign Test

The sign test is a practical method used to determine where a rational expression is positive or negative. It's particularly useful when solving inequalities. After finding the critical points and dividing the number line into corresponding intervals, we pick test points from each interval to evaluate the sign of the expression in that region.

Applying the Sign Test

For our exercise, we chose test points such as \(x=-5\), \(x=0\), \(\frac{3}{2}\), \(\frac{5}{2}\), and \(x=4\) within the intervals defined by the critical points. By plugging these test points into the rational expression, we determined the sign of the expression in each interval. This process is crucial as it directly leads us to the intervals that satisfy the original inequality.

Factorization

Factorization is breaking down a complex expression into a product of simpler factors. In algebra, this often involves transforming polynomials into a product of binomials or other polynomials with lower degrees. Factorization is a key step in solving equations and inequalities because it can significantly simplify expressions and reveal their roots and undefined points.

Benefits of Factorization

For the given problem, factorization allowed us to rewrite the quadratic expressions in the numerator and denominator as \(x^2 + x - 12 = (x + 4)(x - 3)\) and \(x^2 - 3x + 2 = (x - 1)(x - 2)\), respectively. With the rational expression in factored form, identifying the critical points and applying the sign test became more manageable. As such, factorialization is not merely a step; it’s a powerful tool that helps in gaining insights into the properties of algebraic expressions.