Problem 86 - Intermediate Algebra

Question

Solve each equation. See Sections 2.1 and 5.8. $$ (y+2)(5 y+4)=0 $$

Step-by-Step Solution

Answer

The solutions are \(y = -2\) and \(y = -\frac{4}{5}\).

Step 1: Understand the Zero Product Property

The equation \((y+2)(5y+4)=0\) is a product of two expressions set equal to zero. The Zero Product Property states that if the product of two terms is zero, then at least one of the terms must be zero. This means that either \((y+2)=0\) or \((5y+4)=0\).

Step 2: Solve the First Equation

Set the first expression equal to zero: \(y + 2 = 0\). Solve for \(y\) by subtracting 2 from both sides of the equation.\[\begin{align*}y + 2 &= 0 \y &= -2\end{align*}\]

Step 3: Solve the Second Equation

Set the second expression equal to zero: \(5y + 4 = 0\). Solve for \(y\) by first subtracting 4 from both sides, and then dividing by 5.\[\begin{align*}5y + 4 &= 0 \5y &= -4 \y &= -\frac{4}{5}\end{align*}\]

Step 4: State the Solutions

The solutions to the equation \((y+2)(5y+4)=0\) are the values of \(y\) that make either factor equal to zero. From the previous steps, we have determined these values to be \(y = -2\) and \(y = -\frac{4}{5}\).

Key Concepts

Solving Quadratic Equations

Solving quadratic equations involves finding the roots (solutions) of a quadratic equation, which is any equation that can be rearranged to the standard form \(ax^2 + bx + c = 0\). In the given exercise, the equation \((y+2)(5y+4)=0\) is already factored, simplifying the process. Quadratic equations typically have two solutions, corresponding to the values of the variable that make the equation true.
Understanding the Zero Product Property is crucial here. This principle states if the product of two expressions is zero, then at least one of the expressions must be zero. Thus, to solve the quadratic equation, we set each factor equal to zero and solve for the variable.
Quadratic equations can be solved using various methods, including the quadratic formula, completing the square, or graphing. However, when the equation can be factored, it's often the quickest and simplest method.

Algebraic Solutions

Algebraic solutions refer to solving equations using algebraic methods. These involve manipulating the equation through addition, subtraction, multiplication, and division. The goal is to isolate the variable, making it easier to find its value.
In the step-by-step solution for the equation \((y+2)(5y+4)=0\), algebraic methods are applied by setting each factor of the equation to zero. We see the manipulations of equations such as adding or subtracting terms and dividing by coefficients.

• For \(y + 2 = 0\): Subtract 2 from both sides to solve for \(y\).
• For \(5y + 4 = 0\): Subtract 4 from both sides, then divide by 5.

These straightforward operations are fundamental to algebra, embodying the principle that whatever is done to one side of an equation must be done to the other to maintain equality.

Factorization Techniques

Factorization techniques are methods used to rewrite expressions as the product of their simplest terms, or factors. It is a critical step in solving many algebraic equations, especially quadratics.
The equation \((y+2)(5y+4)=0\) has already been factored, showing a product of two linear expressions. Each factor represents a linear equation derived from breaking down the original quadratic equation. The benefit of factorization is that it allows us to apply the Zero Product Property directly.
Effective factorization may involve:

• Identifying common factors among terms.
• Using special product patterns, such as the difference of squares.
• Factoring by grouping, where terms are rearranged and regrouped to find simpler common factors.

Mastering these techniques helps solve complex equations by reducing them to simpler ones, revealing solutions more clearly.