Problem 19 - Basic Technical Mathematics with Calculus

Question

$$\text { Solve the given quadratic equations by factoring.}$$ $$27 m^{2}=3$$

Step-by-Step Solution

Answer

The solutions are \(m = \frac{1}{3}\) and \(m = -\frac{1}{3}\).

Step 1: Move All Terms to One Side

Start solving the equation by moving all terms to one side. Begin with the quadratic equation: \[27m^2 = 3\]Subtract 3 from both sides to set the equation to zero:\[27m^2 - 3 = 0\]

Step 2: Factor the Equation

Next, factor out the greatest common factor from the equation. Notice that both terms in the equation, \(27m^2\) and \(-3\), have a common factor of 3. Factor out 3 from the equation:\[3(9m^2 - 1) = 0\]

Step 3: Identify and Factor the Difference of Squares

The expression inside the parentheses, \(9m^2 - 1\), is a difference of squares. This can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\). Identify \(a\) and \(b\):\[a = 3m, \quad b = 1\]Apply the formula:\[3(3m - 1)(3m + 1) = 0\]

Step 4: Set Each Factor to Zero

To find the solutions for \(m\), set each factor equal to zero and solve for \(m\):1. \(3(3m - 1) = 0\) leads to consequences that all terms involve 0: - \(3m - 1 = 0\) Add 1 to both sides: \[3m = 1\] Divide by 3: \[m = \frac{1}{3}\]2. \(3(3m + 1) = 0\) - \(3m + 1 = 0\) Subtract 1 from both sides: \[3m = -1\] Divide by 3: \[m = -\frac{1}{3}\]

Step 5: Verify the Solutions

Substitute both solutions back into the original equation to ensure they satisfy \(27m^2 = 3\):- For \(m = \frac{1}{3}\): \[27\left(\frac{1}{3}\right)^2 = 3\] \[27\left(\frac{1}{9}\right) = 3\] \[3 = 3\] (True)- For \(m = -\frac{1}{3}\): \[27\left(-\frac{1}{3}\right)^2 = 3\] \[27\left(\frac{1}{9}\right) = 3\] \[3 = 3\] (True)Both values are solutions to the original equation.

Key Concepts

Factoring Quadratic Equations

To solve quadratic equations by factoring, the goal is to express the quadratic equation in the form \( ax^2 + bx + c = 0 \) into a product of factors. This method simplifies finding the values of the unknown variable by breaking down complex expressions into manageable factors.
Factoring involves identifying coefficients, looking for common factors and special patterns within the equation. For instance, the equation \( 27m^2 - 3 = 0 \) can be simplified by factoring out the greatest common factor, which is 3 in this case. This transforms the equation into \( 3(9m^2 - 1) = 0 \).
Simplicity in the form of common factors aids in subsequent steps, such as applying formulas like the difference of squares, enhancing efficiency in finding accurate solutions.

Difference of Squares

The difference of squares is a special technique in factoring quadratic equations. It applies when you have an expression of the form \( a^2 - b^2 \). This expression can be factored into \( (a - b)(a + b) \).
In our case, after factoring out the greatest common factor in \( 9m^2 - 1 \), we recognize it as a difference of squares. Here, \( 9m^2 \) is \((3m)^2\) and \( 1 \) is \(1^2\), making it ideal for this factoring method.
By applying the difference of squares formula, the equation \( 9m^2 - 1 \) becomes \((3m - 1)(3m + 1)\). This process is crucial because it further simplifies the equation, paving the way to solve it efficiently using the zero product property.

Zero Product Property

The zero product property states that if the product of two numbers is zero, then at least one of the numbers must be zero. This principle is particularly useful in solving quadratic equations when they are factored.
Given the factored form \( 3(3m - 1)(3m + 1) = 0 \), we apply the zero product property. Each separate factor is set to zero: \( 3m - 1 = 0 \) and \( 3m + 1 = 0 \), respectively.
Solving these simple linear equations gives the solutions \( m = \frac{1}{3} \) and \( m = -\frac{1}{3} \). Utilizing this property capitalizes on the idea that non-zero factors won't affect zero outcomes, significantly reducing the complexity involved in solving quadratic equations.

Quadratic Equation Verification

Verifying the solutions of quadratic equations ensures that the obtained roots do satisfy the original equation. This step reconfirms the correctness of the entire solving process.
Once \( m = \frac{1}{3} \) and \( m = -\frac{1}{3} \) are identified as potential solutions, they must be substituted back into the equation \( 27m^2 = 3 \). Performing these substitutions checks the accuracy:

• For \( m = \frac{1}{3} \), substituting gives \( 27\left(\frac{1}{3}\right)^2 = 3 \), simplifying to \( 3 = 3 \).
• For \( m = -\frac{1}{3} \), substituting gives \( 27\left(-\frac{1}{3}\right)^2 = 3 \), leading again to \( 3 = 3 \).

Both verify as true, reassuring that the solutions are correct. Verification is indispensable as it ties back the mathematical process to the original problem, confirming the equation's integrity.