Problem 71
Question
Why is the rational expression \(\frac{x+3}{x^{2}-4}\) undefined for \(x=2\) and \(x=-2\) but defined for \(x=-3 ?\)
Step-by-Step Solution
Verified Answer
The expression is undefined when the denominator is zero (at \(x=2\) and \(x=-2\)), defined when it is non-zero (at \(x=-3\)).
1Step 1: Identify Conditions for Undefined Expressions
A rational expression is undefined when its denominator equals zero. Therefore, identify the values of \(x\) that make the denominator zero.
2Step 2: Set Denominator to Zero and Solve
Given the expression \(\frac{x+3}{x^{2}-4}\), the denominator is \(x^{2}-4\). Set this equal to zero: \(x^{2}-4=0\).
3Step 3: Solve the Quadratic Equation
Solve the equation \(x^{2}-4=0\) for \(x\): Factor it as \((x-2)(x+2)=0\). Thus, \(x=2\) and \(x=-2\) are the solutions.
4Step 4: State the Undefined Conditions
The expression is undefined for \(x=2\) and \(x=-2\) because those values make the denominator zero.
5Step 5: Evaluate at \(x=-3\)
Evaluate the expression when \(x=-3\): The denominator becomes \((-3)^{2}-4=9-4=5\), which is non-zero.
6Step 6: Conclusion on Defined or Undefined Status
Since the denominator is non-zero when \(x=-3\), the expression is defined at this point.
Key Concepts
Undefined ExpressionsQuadratic EquationsFactoring Polynomials
Undefined Expressions
Rational expressions can sometimes be tricky, especially when they become undefined. An undefined expression occurs when the denominator of a rational expression equals zero. Why is this important? Because any number divided by zero is undefined in mathematics! In our example, we are given the rational expression \(\frac{x+3}{x^2-4}\). To find when it's undefined, we set the denominator equal to zero: \(x^2-4=0\). When we solve this, we find that the expression is undefined for \(x=2\) and \(x=-2\) because these values make the denominator zero and, therefore, the entire expression undefined. It is key to remember:
- Undefined expressions occur when the denominator is zero.
- Setting the denominator to zero helps identify problematic values of \(x\).
- Rational expressions are only defined when the denominator is non-zero.
Quadratic Equations
Quadratic equations are invaluable tools in algebra. They usually appear in the form \(ax^2 + bx + c = 0\). Solving these equations often helps determine when an expression, like our rational one, is undefined. In the problem, you encounter the equation \(x^2-4=0\). This is a simplified quadratic equation with no \(b\) term. Here's how you solve it:
- Identify the equation to solve: \(x^2-4=0\).
- Rearrange if needed, but here it is ready to go.
- Solve by factoring, completing the square, or using the quadratic formula.
- Simplifying quadratic equations can reveal when expressions are undefined.
- Factoring is a common method to solve simple quadratics.
Factoring Polynomials
Factoring polynomials is a crucial skill in tackling algebraic problems, including identifying when expressions are undefined. It's the process of breaking down a polynomial into simpler factors that, when multiplied together, give you the original polynomial. In the equation \(x^2-4\), we factor it as \((x-2)(x+2)\), which effortlessly shows us the values of \(x\) that result in the polynomial equating to zero.
Key points to master factoring include:
Key points to master factoring include:
- Recognize patterns – some quadratics can fall into known identities like difference of squares.
- Break down the polynomial using these patterns, if possible.
- Test your factors by multiplying them back to see if they reconstruct the original polynomial.
Other exercises in this chapter
Problem 70
What role does factoring play in the simplifying of rational expressions?
View solution Problem 71
For which real numbers does \(\frac{x}{x-3}+\frac{4}{x}\) equal \(\frac{(x+6)(x-2)}{x(x-3)} ?\) Explain your answer.
View solution Problem 72
Suppose that your friend does an addition problem as follows: $$ \frac{5}{8}+\frac{7}{12}=\frac{5(12)+8(7)}{8(12)}=\frac{60+56}{96}=\frac{116}{96}=\frac{29}{24}
View solution Problem 72
How would you convince someone that \(\frac{x-4}{4-x}=-1\) for all real numbers except 4 ?
View solution