Problem 15 - Algebra for College Students

Question

Which real numbers cannot be used in place of the variable in each rational expression? $$\frac{2 a-3}{a^{2}+5 a+6}$$

Step-by-Step Solution

Answer

The real numbers that cannot be used are \(a = -2\) and \(a = -3\).

Step 1: Identify the Denominator

The denominator of the expression is the part that can cause the rational expression to be undefined. In this expression, the denominator is \(a^2 + 5a + 6\).

Step 2: Set the Denominator Equal to Zero

To find the values of \(a\) that make the denominator equal to zero, set the equation \(a^2 + 5a + 6 = 0\).

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \(a^2 + 5a + 6 = 0\) by factoring. The equation factors into \((a + 2)(a + 3) = 0\).

Step 4: Find the Solutions

Set each factor equal to zero: \(a + 2 = 0\) and \(a + 3 = 0\). Solving these gives \(a = -2\) and \(a = -3\).

Step 5: Identify Excluded Values

The values \(a = -2\) and \(a = -3\) make the denominator zero and thus the rational expression undefined. Therefore, the real numbers that cannot be used for the variable in this expression are \(-2\) and \(-3\).

Key Concepts

Denominator in Rational Expressions

The denominator in a rational expression plays a crucial role in determining if the expression is defined. If the denominator is zero, the expression becomes undefined.
In our example, the denominator is given by the polynomial equation: \( a^2 + 5a + 6 \) It's important to identify the values of \( a \) that will make this polynomial equal to zero because these values will make the entire expression undefined.

Undefined Expressions

A rational expression becomes undefined when its denominator is zero. To determine the excluded values, you need to solve the equation where the denominator is set to zero.
For the rational expression: \( \frac{2a-3}{a^2+5a+6} \) The denominator \( a^2 + 5a + 6 \) should be set equal to zero: \[ a^2 + 5a + 6 = 0 \] Solving this equation will give the values of \( a \) that make the expression undefined.

Solving Quadratic Equations

To solve the quadratic equation \( a^2 + 5a + 6 = 0 \), you can use the factoring method. Factoring quadratics involves breaking down the quadratic polynomial into two binomials.
In this case, the quadratic equation factors into: \[ (a + 2)(a + 3) = 0 \] Setting each factor equal to zero gives the solutions: \[ a + 2 = 0 \] and \[ a + 3 = 0 \] These equations are then solved for \ a \: Solving the first gives: \[ a = -2 \] and the second: \[ a = -3 \]

Factoring Quadratics

Factoring quadratics is an efficient method to solve quadratic equations, especially when the quadratic can be easily broken into simpler binomial factors.
For \( a^2 + 5a + 6 \), you find two numbers that multiply to give the constant term (6) and add to give the linear coefficient (5). Here, 2 and 3 fit the criteria because:

• 2 * 3 = 6
• 2 + 3 = 5

Therefore, \( a^2 + 5a + 6 \) can be factored as \( (a + 2)(a + 3) \) Using this factored form, solving for when the product equals zero gives you the values \( a = -2 \) and \( a = -3 \)