Problem 85 - Prealgebra and Introductory Algebra

Question

Name the values of \(x\) for which the rational expression is undefined. (Hint: Set the denominator equal to zero and solve for \(x\).) $$\frac{x+5}{x^{2}-4 x-5}$$

Step-by-Step Solution

Answer

The values of \(x\) for which the rational expression is undefined are \(x=5\) and \(x=-1\).

Step 1: Set the Denominator Equal to Zero

We start by setting the denominator of the rational expression equal to zero. \[x^{2}-4 x-5 = 0\]

Step 2: Factor the Quadratic Equation

We then factor the quadratic equation. The factors of -5 that sum up to -4 are -5 and 1. Therefore, we write: \[(x-5)(x+1) = 0\]

Step 3: Solve for \(x\)

Next, set each factor equal to zero and solve for \(x\): \[x-5=0 \Rightarrow x=5\] \[x+1=0 \Rightarrow x=-1\]

Step 4: Check the Solution

To check the solution into the original expression, if the denominator becomes zero, then that is the value of \(x\) we are looking for. Checking for \(x=5\) and \(x=-1\), we find that the denominator becomes zero, hence, our solutions are correct.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental algebraic expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The solutions to these equations represent the points where the graph of the quadratic function intersects the x-axis. These solutions, or roots, can be found using various methods, including factoring, completing the square, and the quadratic formula.

In this exercise, we deal with a quadratic equation found in the denominator \(x^2-4x-5=0\). Notice, it's in its standard form with \(a=1\), \(b=-4\), and \(c=-5\). The goal is to find the values of \(x\) that satisfy this equation, as these make the original rational expression undefined.

Identifying where the quadratic equation equals zero helps determine where the denominator becomes zero, creating undefined values for the rational expression.

Factoring

Factoring is a technique used to break down quadratic equations into simpler expressions that can be multiplied together to give the original equation. The core idea is to find two numbers that multiply to the constant term \(c\) and add up to the linear coefficient \(b\).

In this case, we need to factor \(x^2-4x-5\). We look for two numbers that multiply to \(-5\) (constant term) and add to \(-4\) (linear coefficient). These numbers are \(-5\) and \(+1\). Thus, the equation \(x^2-4x-5\) factors into \((x-5)(x+1) = 0\).

Once factored, we set each factor equal to zero and solve for \(x\). This gives us the solutions \(x=5\) and \(x=-1\). Factoring offers a straightforward method to find the roots of simple quadratic equations.

Undefined Expressions

Rational expressions are undefined when their denominators equal zero. This is because division by zero is an operation that doesn’t result in a finite number, leading to an undefined expression. To address this, we need to determine the \(x\) values that make the denominator zero.

In our exercise, we found the denominator \(x^2-4x-5\) factors to \((x-5)(x+1)\). Using these factors, setting each to zero provides \(x=5\) and \(x=-1\). These are the \(x\) values that make the denominator zero. Thus, the rational expression \(\frac{x+5}{x^2-4x-5}\) is undefined at these points.

Checking these values by substituting them back into the denominator confirms that they indeed make the denominator zero, confirming their status as undefined values for the expression.