Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Show that the equation is an identity. $$\frac{x^{2}-9}{x+3}=x-3$$

Step-by-Step Solution

Answer

The equation \( \frac{x^{2}-9}{x+3}=x-3 \) is an identity for all \( x \neq -3 \).

Step 1: Understand the Components of the Equation

The given equation is \( \frac{x^{2}-9}{x+3}=x-3 \). An identity is an equation that holds true for all values of the variable for which both sides of the equation are defined.

Step 2: Factor the Numerator

The numerator \( x^2 - 9 \) is a difference of squares, which can be factored using the formula \( a^2 - b^2 = (a+b)(a-b) \). Here, \( a = x \) and \( b = 3 \). So, \( x^2 - 9 = (x+3)(x-3) \).

Step 3: Simplify the Fraction

With the factored form of the numerator, the equation becomes \( \frac{(x+3)(x-3)}{x+3} \). We can simplify this by canceling the \( x+3 \) terms. However, this is only valid if \( x+3 eq 0 \) (i.e., \( x eq -3 \)), to avoid division by zero.

Step 4: Result After Simplification

After canceling \( x+3 \), the fraction simplifies to \( x-3 \). Thus the equation \( \frac{(x+3)(x-3)}{x+3} = x-3 \) is equivalent to \( x-3 = x-3 \) whenever \( x eq -3 \).

Step 5: Verify Identity Conditions

Since \( x-3 = x-3 \) for all \( x eq -3 \), this equation holds true for all defined values of \( x \). Therefore, the original equation is an identity, because it simplifies to a true statement given the domain exclusion.

Key Concepts

Difference of Squares

The difference of squares is a unique algebraic identity useful for factoring polynomials where subtraction is involved. In general, the formula looks like this:

• For any two terms, say, \( a \) and \( b \), the difference of their squares can be expressed as \( a^2 - b^2 = (a + b)(a - b) \).

The simplicity and symmetry of this identity make it a powerful tool in algebra.
Let's break it down with the given expression \( x^2 - 9 \).
We can see that \( x^2 \) is a square and 9 is a square number, being \( 3^2 \). This implies we can rewrite it as:

• \( x^2 - 9 = (x)^2 - (3)^2 \)
• According to the difference of squares formula, it factors to \( (x + 3)(x - 3) \).

This identity simplifies complex quadratic expressions and assists in breaking them down into products of linear terms.

Rational Expressions

Rational expressions in mathematics are similar to rational numbers. They are fractions where the numerator and the denominator are polynomials. When working with rational expressions, one crucial rule is to simplify whenever possible, keeping in mind division by zero constraints.
In the original exercise, the equation given is \( \frac{x^2 - 9}{x + 3} = x - 3 \). After factoring the numerator as \( (x + 3)(x - 3) \), the expression becomes a rational one:

• The expression \( \frac{(x + 3)(x - 3)}{x + 3} \) suggests cancelling out the common \( x + 3 \) terms.

This step results in \( x - 3 \) when \( x + 3 \) is not equal to zero, ensuring no undefined expressions due to division by zero.
Paying attention to restrictions and simplifying them is key for correct and valid manipulations of rational expressions.

Factoring

Factoring is the process of writing a polynomial or algebra expression as a product of simpler terms. This skill is crucial when simplifying expressions or solving equations.
For example, in the exercise provided, the given polynomial \( x^2 - 9 \) can be factored using the difference of squares identity into \( (x + 3)(x - 3) \). This conversion to a product format allows us to simplify or solve the expression more easily.

• Factoring the expression makes it simpler to cancel common terms with the denominator if possible.
• This process also gives insight into the roots or zeros of the polynomial, if solving an equation.

It's important to remember that not all polynomials can be factored over the rational numbers easily, but techniques like the difference of squares play an essential role in the factoring process.
Embracing factoring as a fundamental skill will enhance your ability to tackle a wide range of algebraic problems confidently.