Problem 32 - Intermediate Algebra

Question

Solve each equation. $$ \frac{3}{t+2}+\frac{4}{t-2}=2 $$

Step-by-Step Solution

Answer

The solutions are \( t = \frac{7 \pm \sqrt{129}}{4} \).

Step 1: Find a Common Denominator

To solve the equation \( \frac{3}{t+2} + \frac{4}{t-2} = 2 \), first identify a common denominator for the fractions. The common denominator of \((t+2)\) and \((t-2)\) is \((t+2)(t-2)\).

Step 2: Clear the Fractions

Multiply every term of the equation by the common denominator \((t+2)(t-2)\) to eliminate the fractions:\[ (t+2)(t-2) \cdot \frac{3}{t+2} + (t+2)(t-2) \cdot \frac{4}{t-2} = 2 \cdot (t+2)(t-2) \]This simplifies to:\[ 3(t-2) + 4(t+2) = 2(t+2)(t-2) \].

Step 3: Expand and Simplify the Equation

Expand the terms on both sides:\[ 3t - 6 + 4t + 8 = 2(t^2 - 4) \]Combine like terms:\[ 7t + 2 = 2t^2 - 8 \].

Step 4: Rearrange into Quadratic Form

Rearrange the equation to form a standard quadratic equation:\[ 2t^2 - 7t - 10 = 0 \].

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2 \), \( b = -7 \), \( c = -10 \):\[ t = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 2 \times (-10)}}{4} \]\[ = \frac{7 \pm \sqrt{49 + 80}}{4} \]\[ = \frac{7 \pm \sqrt{129}}{4} \].

Step 6: Determine Valid Solutions

After evaluating, we have two potential solutions:\[ t = \frac{7 + \sqrt{129}}{4} \quad \text{and} \quad t = \frac{7 - \sqrt{129}}{4} \].Check if any solutions cause the original denominators to be zero. Neither value causes a zero denominator in the original fractions, so both solutions are valid.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides the solution for \( x \) using the coefficients \( a \), \( b \), and \( c \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The symbol \( \pm \) indicates that there are typically two solutions, called roots, to the equation. This formula is useful because it finds the exact solutions even if the roots are complex numbers.

• Identify \( a \), \( b \), and \( c \) from your equation.
• Plug these values into the quadratic formula.
• Simplify under the square root and solve for \( x \).

In our exercise, the quadratic formula helps us determine the potential solutions for \( t \) once we rearrange the equation into the traditional quadratic format. Understanding this concept deeply allows students to manage equations that are often seen as challenging easily.

Common Denominator

Creating a common denominator is essential when dealing with fractions, particularly in rational equations. A common denominator is simply a shared multiple of the denominators of the fractions you are working with. This allows you to combine or manipulate the fractions as needed.In our original exercise, the fractions \( \frac{3}{t+2} \) and \( \frac{4}{t-2} \) have different denominators. To eliminate the fractions and simplify the equation, it's crucial to find a common denominator:

• The common denominator for \( t+2 \) and \( t-2 \) is \((t+2)(t-2)\).

Use this shared denominator to clear the fractions by multiplying each term by the common denominator. This process helps us transform the equation into a more manageable form, free from fractions, making it easier to proceed to find the solution.

Expanding Expressions

Expanding expressions involves using algebraic techniques to simplify equations, particularly when dealing with products of binomials or polynomials. This process makes complex expressions more straightforward to work with and helps to identify like terms to combine.In the step-by-step solution to our exercise, we come across terms such as:

• \(3(t-2)\)
• \(4(t+2)\)
• \(2(t^2 - 4)\)

Expanding these expressions involves distributing the multiplication over each term within the brackets:

• \(3(t-2) = 3t - 6\)
• \(4(t+2) = 4t + 8\)
• \(2(t^2 - 4) = 2t^2 - 8\)

Combine like terms after expanding to further simplify the expression, paving the way for easier manipulation towards finding the solution.

Validation of Solutions

Once potential solutions to the equation are found, validating these solutions is crucial. This ensures that the solutions do not cause any original denominators to be zero, as that would make the equation undefined.In the exercise at hand, we discovered potential solutions for \( t \):

• \( t = \frac{7 + \sqrt{129}}{4} \)
• \( t = \frac{7 - \sqrt{129}}{4} \)

To validate these solutions, substitute them back into the original denominators \((t+2)\) and \((t-2)\) to verify that neither causes a division by zero.

• If any solution causes a zero denominator, discard it as invalid.
• In this exercise, both solutions are valid since they don't zero out the denominators.

This step ensures the reliability of the solutions, a key checkpoint before considering the solutions correct and final.