Problem 14 - Elementary Algebra

Question

For the following problems, solve the equations. $$ 6 r^{2}-36=0 $$

Step-by-Step Solution

Answer

Answer: The solutions for the quadratic equation are r = -2 and r = 3.

Step 1: Identify the type of equation

This is a quadratic equation in the form of $$ar^2 + br + c = 0$$, where a = 6, b = 0, and c = -36.

Step 2: Factor the equation

We will factor the quadratic equation to find its solutions. Since there are no terms with r (degree 1), we only need to factor the terms with r^2 (degree 2) and the constant term. Look for two numbers whose product is -216 (6*(-36)) and whose sum is 0. Those numbers are 18 and -12. $$ \begin{aligned} 6r^2 - 36 &= 0 \\ 6(r^2 - 6) &= 0 \\ 6(r + 2)(r - 3) &= 0 \end{aligned} $$

Step 3: Find the solutions for r

Now that we have factored the equation, we can find the solutions for r by setting each factor equal to zero and solving for r. $$ \begin{aligned} r + 2 &= 0 \\ r &= -2 \end{aligned} $$ $$begin{aligned} r - 3 &= 0 \\ r &= 3 \end{aligned} $$ So, the solutions for the equation are r = -2 and r = 3.

Key Concepts

Factoring

Factoring is an essential skill in algebra, especially when dealing with quadratic equations. It involves expressing a polynomial as a product of two or more simpler expressions, called factors. This can help in simplifying an equation and making it easier to solve. In our original exercise, we have the quadratic equation \(6r^2 - 36 = 0\). Here’s how to factor it:

• First, note the equation is in the form \(ar^2 + br + c = 0\), where \(a = 6\), \(b = 0\), and \(c = -36\).
• Since there is no \(r\) term (\(b = 0\)), focus on \(r^2\) and the constant \(c\).
• Next, identify two numbers that multiply to \(6 \times -36 = -216\) and add up to \(0\). These are \(18\) and \(-12\).
• Use these numbers to rewrite the equation as \(6(r + 2)(r - 3) = 0\).

Through factoring, we can transform the equation into a simpler product form. This allows us to identify potential solutions by looking at each factor individually.

Solutions of Equations

Finding solutions to equations is the ultimate goal when solving them, as it provides the specific values that make the equation true. In the context of our quadratic equation \(6r^2 - 36 = 0\), we arrived at a factored form: \(6(r + 2)(r - 3) = 0\). Here's how we find the solutions:

• Set each factor equal to zero: \(r + 2 = 0\) and \(r - 3 = 0\).
• Solve each simple equation. For \(r + 2 = 0\), subtract \(2\) from both sides to get \(r = -2\).
• For \(r - 3 = 0\), add \(3\) to both sides, yielding \(r = 3\).

These solutions, \(r = -2\) and \(r = 3\), are the values of \(r\) that satisfy the original equation. By checking these solutions, we can ensure they make the initial equation equal to zero, hence proving their validity.

Algebraic Expressions

Algebraic expressions form the backbone of algebra and are crucial in forming and solving equations. They consist of variables, coefficients, and constants combined through operations like addition, subtraction, multiplication, and division. Our original problem involves an algebraic expression within a quadratic equation: \(6r^2 - 36\).Breaking it down:

• Variables like \(r\) are symbols that represent unknown values. In quadratic equations, they often appear squared, as seen with \(r^2\).
• Coefficients are the numbers multiplying the variables, such as \(6\) in \(6r^2\). They help determine the expression's overall scale and impact.
• Constants, like \(-36\), are fixed values that don't change. They define the equation's shift along the y-axis on a graph.

Understanding these components is crucial for manipulating expressions algebraically, allowing you to factor, expand, and simplify to solve equations correctly. Mastery of these concepts makes dealing with complex equations more approachable.