Problem 22 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$9=a^{2} x^{2}$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{3}{a} \) and \( x = \frac{-3}{a} \).

Step 1: Rewrite the Equation

The original equation is \( 9 = a^{2}x^{2} \). Begin by rewriting the equation such that one side equals zero, resulting in \( a^{2}x^{2} - 9 = 0 \).

Step 2: Recognize the Perfect Square

Notice that the equation \( a^{2}x^{2} - 9 = 0 \) is a difference of squares. You can recognize this because both \( a^{2}x^{2} \) and 9 are perfect squares.

Step 3: Apply the Difference of Squares Formula

The difference of squares formula is \( p^2 - q^2 = (p-q)(p+q) \). Here, \( p = ax \) and \( q = 3 \), thus the equation can be factored as \( (ax - 3)(ax + 3) = 0 \).

Step 4: Set Each Factor to Zero

To solve for \( x \), set each factor equal to zero: \( ax - 3 = 0 \) and \( ax + 3 = 0 \).

Step 5: Solve Each Equation

Solve for \( x \) in each equation. \( ax - 3 = 0 \) gives \( ax = 3 \), so \( x = \frac{3}{a} \). Similarly, \( ax + 3 = 0 \) gives \( ax = -3 \), so \( x = \frac{-3}{a} \).

Step 6: Combine Solutions

The solutions for the equation \( a^{2}x^{2} = 9 \) are \( x = \frac{3}{a} \) and \( x = \frac{-3}{a} \).

Key Concepts

Factoring

Factoring is a pivotal method for solving quadratic equations.
This involves expressing a quadratic expression as a product of simpler expressions.
In this context, specifically, we're dealing with an equation like \( a^{2}x^{2} - 9 = 0 \).
Generally, the first step with such equations is to arrange all components of the equation so one side equals zero if it isn't already set up this way.

• Adjust your equation: Get everything on one side to formulate \( a^{2}x^{2} - 9 \), as seen in our example.
• Look for patterns: A prime example is recognizing perfect square terms, making it easier to factor.

After the arrangement, one can use known formulas and techniques, such as the difference of squares, to break down the expression into products of simpler terms.
Factoring aims to transform the equation to an equivalent set of simpler expressions, which can then be solved individually.

Difference of Squares

Understanding the difference of squares is crucial when factoring specific kinds of quadratic expressions.
The difference of squares is a mathematical pattern: \( p^2 - q^2 \), where both \( p \) and \( q \) are perfect squares.
This formula simplifies as follows: \( (p-q)(p+q) \).
Identifying a difference of squares situation often makes factoring quick and straightforward.

• Recognize the Pattern: Both \( a^{2}x^{2} \) and \( 9 \) are perfect squares. Hence, it suggests using this handy formula.
• Application: Set \( p = ax \) and \( q = 3 \) in your equation to apply the difference of squares, which works effectively as \( (ax-3)(ax+3) \).

Using the difference of squares can be a great shortcut for factoring complex quadratic equations, making them more manageable and easier to solve.

Solution Methods

Once your quadratic equation is factored, you have two expressions set to zero, thanks to factoring techniques like the difference of squares.
These simpler equations can now be solved individually. This involves applying the zero-product property, which states if \( xy = 0 \), then \( x = 0 \) or \( y = 0 \).
This principle allows you to isolate each factor and solve for the variable independently.

• Set Each Factor to Zero: For example, with factors \( (ax-3) \) and \( (ax+3) \), we solve:
  • \( ax-3=0 \) gives \( ax=3 \), leading to \( x=\frac{3}{a} \).
  • \( ax+3=0 \) yields \( ax=-3 \), resulting in \( x=\frac{-3}{a} \).
• Combine the Solutions: After solving each, gather all solutions to identify the complete set of roots for the quadratic equation.

This method goes a long way in equipping one to solve various kinds of quadratic equations efficiently and effectively.