Problem 20 - Basic Technical Mathematics with Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Move All Terms to One Side

Start by moving all terms of the equation to one side of the equation to set it equal to zero. The original equation is already set with zero on one side: \[ a^2 x^2 - 9 = 0 \]

Step 2: Recognize as a Difference of Squares

Notice that the equation \( a^2 x^2 - 9 = 0 \) is a difference of squares. A difference of squares is a term in the form \( A^2 - B^2 \), which can be factored into \( (A-B)(A+B) \). Here, \( A = ax \) and \( B = 3 \), so: \[ a^2 x^2 - 9 = (ax)^2 - 3^2 \]

Step 3: Factor the Equation

Apply the difference of squares formula to factor the equation: \[ (ax - 3)(ax + 3) = 0 \]

Step 4: Set Each Factor to Zero

To find the solutions, set each factor equal to zero and solve for \( x \): 1. \( ax - 3 = 0 \) 2. \( ax + 3 = 0 \)

Step 5: Solve Each Equation

Solve each equation for \( x \): 1. For \( ax - 3 = 0 \): \[ ax = 3 \] \[ x = \frac{3}{a} \] (as long as \( a eq 0 \))2. For \( ax + 3 = 0 \): \[ ax = -3 \] \[ x = \frac{-3}{a} \] (as long as \( a eq 0 \))

Step 6: State the Solution

The solutions for the quadratic equation \( 9 = a^2 x^2 \) are:\[ x = \frac{3}{a} \quad \text{and} \quad x = \frac{-3}{a} \]

Key Concepts

Factoring Quadratics

When we talk about factoring quadratics, we're diving into a method used to simplify equations and solve for unknown values. Quadratic equations are usually in the form \( ax^2 + bx + c = 0 \). Our goal when factoring is to express this quadratic as a product of two binomial expressions, which makes it easier for us to solve.

• The key is to find numbers that multiply to give you \( ac \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( b \) (the coefficient of \( x \)).
• Once these numbers are identified, we can break down the middle term and factor by grouping.

In the specific equation \( 9 = a^2 x^2 \), the factorization process is simplified by recognizing it as a type known as a difference of squares (more on that next!). Factoring quadratics is a crucial skill that allows us to solve equations easily by turning them into simpler, more manageable forms.

Difference of Squares

The difference of squares is a special pattern in mathematics where two perfect squares are subtracted. This pattern can be identified and factored quickly, which is very helpful when working with quadratic equations. The general formula for difference of squares is \( A^2 - B^2 = (A-B)(A+B) \).

• In our given exercise, the equation \( a^2 x^2 - 9 = 0 \) is clearly a difference of squares because \( 9 \) is a perfect square \( (3^2) \) and \( a^2 x^2 \) is also a perfect square \( ((ax)^2) \).
• By applying the difference of squares formula, we break this into \( (ax - 3)(ax + 3) = 0 \).

Recognizing this pattern allows us to factor the equation quickly and, consequently, leads to the solutions. Always be on the lookout for this handy pattern when working with quadratics!

Solving Equations

Once a quadratic equation is factored, solving it becomes a straightforward task. This involves finding the values that make the product of factors zero. According to the zero product property, if the product of two factors is zero, at least one of the factors must be zero.

• For \( (ax - 3)(ax + 3) = 0 \), set each factor equal to zero: \( ax - 3 = 0 \) and \( ax + 3 = 0 \).
• Solve each equation individually: \( ax = 3 \) or \( ax = -3 \).
• Divide by \( a \) (provided \( a eq 0 \)) to isolate \( x \). This yields solutions \( x = \frac{3}{a} \) and \( x = \frac{-3}{a} \).

By solving each equation derived from our factors, we've determined the possible solutions for \( x \). This systematic approach is effective and efficient for solving quadratic equations, making it a fundamental skill in algebra.