Problem 99 - Algebra 1

Question

For what values of the variable is the rational expression undefined? $$\frac{x+4}{x^{2}+x-6}$$

Step-by-Step Solution

Answer

The rational expression is undefined when \(x = 2\) or \(x = -3\).

Step 1: Identify the Denominator

The denominator of the given rational expression is \(x^{2} + x - 6\). This is the part of the expression we are interested in, because if it equals zero, the whole expression becomes undefined.

Step 2: Equate Denominator to Zero and Solve

Now, to find the values for which the denominator equals zero, set the denominator equal to zero and solve for \(x\). \n So, \(x^{2} + x - 6 = 0\). This is a quadratic equation, for which solutions can be found either by factoring, completing the square, or using the quadratic formula. Here it can be factored into \((x-2)(x+3) = 0\). By setting each factor equal to zero and solving, we get \(x = 2\) and \(x = -3\)

Step 3: Summarize the Results

The rational expression is undefined for \(x = 2\) and \(x = -3\). These are the values that make the denominator zero, leading to division by zero, which is undefined.

Key Concepts

Undefined Values

Rational expressions are similar to fractions, consisting of a numerator and a denominator. However, unlike simple fractions, rational expressions can sometimes be undefined. This occurs when the denominator equals zero. Just as you cannot divide by zero in arithmetic, in algebra, a division by zero is "undefined." To find these critical points, we equate the denominator to zero and solve for the variable involved.
For example, in the expression \( \frac{x+4}{x^{2}+x-6} \), to find the undefined values, focus on the denominator \( x^2 + x - 6 \). By solving \( x^2 + x - 6 = 0 \), we discover which values of \( x \) make the denominator zero, thereby making the entire expression undefined.

• Set the denominator equal to zero.
• Solve the resulting equation.
• The solutions indicate when the expression is undefined.

Quadratic Equations

A quadratic equation is any equation that can be rearranged in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations are fundamental in algebra and appear frequently in different types of problems, including when solving undefined values in rational expressions.
Quadratic equations can be solved using several methods, each useful depending on the situation:

• Factoring: when the quadratic expression can be written as a product of two binomials.
• Completing the square: transforming the equation into a perfect square trinomial.
• Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides a universal method for finding solutions.

In the rational expression \( \frac{x+4}{x^2+x-6} \), the denominator \( x^2+x-6 \) is a quadratic that can be solved by factoring. Solving the equation gives us the critical points where the expression becomes undefined.

Factoring Quadratics

Factoring quadratics is a technique we use when solving equations like \( x^2 + x - 6 = 0 \). Factoring involves expressing the quadratic as a product of two binomial expressions.
For instance, the quadratic \( x^2 + x - 6 \) can be factored into \((x-2)(x+3) = 0\). By expressing it in this factored form, we can quickly find the solutions by applying the Zero Product Property, which states that if a product of factors equals zero, at least one of the factors must be zero.
Let's break down the steps:

• Look for two numbers that multiply to \( c \) (constant term in \( ax^2+bx+c\)) and add to \( b\) (linear coefficient).
• Write the quadratic as \((x - p)(x + q)\), where \( p \) and \( q \) are the numbers found.
• Set each factor equal to zero: \( x - p = 0 \) and \( x + q = 0 \).
• Solve for \( x \) to find the roots.

In our example, solving \((x-2)(x+3) = 0\) gives us \( x = 2 \) and \( x = -3 \). These roots identify the values where the expression is undefined.