Problem 16 - Beginning and Intermediate Algebra

Question

Solve each equation.. $$\left(t-\frac{9}{8}\right)\left(t+\frac{5}{6}\right)=0$$

Step-by-Step Solution

Answer

The solutions for the given equation \(\left(t-\frac{9}{8}\right)\left(t+\frac{5}{6}\right)=0\) are \(t = \frac{9}{8}\) and \(t = -\frac{5}{6}\).

Step 1: Set the first factor equal to zero

Set the first factor equal to zero: \(t - \frac{9}{8} = 0\)

Step 2: Solve for t in the first equation

Add \(\frac{9}{8}\) to both sides of the equation: \(t = \frac{9}{8}\)

Step 3: Set the second factor equal to zero

Now, set the second factor equal to zero: \(t + \frac{5}{6} = 0\)

Step 4: Solve for t in the second equation

Subtract \(\frac{5}{6}\) from both sides of the equation: \(t = -\frac{5}{6}\)

Step 5: Combine the solutions

Now we have found both solutions for t: \(t = \frac{9}{8}\) and \(t = -\frac{5}{6}\)

Key Concepts

Solving Equations

When it comes to solving equations, especially quadratic equations, we are often looking for the values of the unknown variable that make the equation true. In the context of the quadratic equation given, it is expressed in a factored form, i.e., \( \left(t-\frac{9}{8}\right)\left(t+\frac{5}{6}\right)=0 \).
By identifying each part and setting them to zero separately, we can systematically find the solutions for the variable.In other words:

• Set each factor equal to zero.
• Solve the resulting simple linear equations.

This approach helps in finding multiple solutions quickly, as quadratic equations can often have more than one value that satisfies them.

Factoring

Factoring is a crucial algebraic skill used to solve quadratic equations. It involves breaking down an equation into simpler terms that can be multiplied together to result in the original equation. Consider the factorized form we have: \( \left(t-\frac{9}{8}\right)\left(t+\frac{5}{6}\right)=0 \).
This indicates that the quadratic equation in standard form has been split into two binomial expressions.Factoring effectively reduces complexity by:

• Converting a complex quadratic equation into simpler binomials.
• Helping us apply the zero product property efficiently.

Once you recognize an equation is factorable, it simplifies the process of finding roots, as opposed to using more complex methods like completing the square or using the quadratic formula.

Zero Product Property

The zero product property is a fundamental theorem used in algebra to efficiently solve equations set to zero. It states that if the product of two numbers is zero, then at least one of the numbers must be zero.
This means, in our equation, \( \left(t-\frac{9}{8}\right)\left(t+\frac{5}{6}\right)=0 \), either \( t-\frac{9}{8}=0 \) or \( t+\frac{5}{6}=0 \) (or both) must be true for the equation to hold.This property is powerful in equations because:

• It allows us to separate the equation into smaller, easier-to-solve parts.
• Every factor having a zero solution provides multiple potential solutions for the entire equation.

Effectively, the zero product property dictates that if you can factorize an expression leading to zero, you can directly extract solutions from each factor.

Steps in Algebra

Working through algebraic problems systematically often involves a series of clear steps. For quadratic equations, especially when presented in a factored form, follow these general steps:1. **Identify** the equation type: Notice if it can be factorized and recognize its structure.2. **Apply** the Zero Product Property: Set each factor to zero, - For example, with \( t-\frac{9}{8}=0 \) and \( t+\frac{5}{6}=0 \).3. **Solve** each resulting equation separately: - For \( t-\frac{9}{8}=0 \), solve to get \( t=\frac{9}{8} \). - For \( t+\frac{5}{6}=0 \), solve to get \( t=-\frac{5}{6} \).4. **Combine** solutions: List out all potential solutions obtained from each factor.Following these steps ensures a structured approach, minimizing errors and improving understanding of the method to solve quadratic equations effectively.