Problem 42 - Beginning and Intermediate Algebra

Question

Solve each equation. $$54=w(15-w)$$

Step-by-Step Solution

Answer

The solutions for the equation \(54 = w(15-w)\) are \(w = 9\) and \(w = 6\).

Step 1: Distribute w on the right side

Distribute w to both terms inside the parentheses on the right side of the equation: \(54 = w \cdot 15 - w^2 \)

Step 2: Rearrange the terms to create a quadratic equation

Rearrange the terms to create a quadratic equation in the form of \(ax^2 + bx + c = 0\): \(w^2 - 15w + 54 = 0\)

Step 3: Factor the quadratic equation

Factor the quadratic equation: \((w - 9)(w - 6) = 0\)

Step 4: Solve for w

Since the product of two terms is equal to zero, one of them must be zero. So, set each term equal to zero and solve for w: 1. \(w - 9 = 0 \Rightarrow w = 9\) 2. \(w - 6 = 0 \Rightarrow w = 6\) Our solutions for the equation are \(w = 9\) and \(w = 6\).

Key Concepts

Factoring

When solving quadratic equations, factoring is often a go-to method for finding solutions. In simple terms, factoring involves expressing a polynomial as a product of its roots or simpler polynomials. Imagine you have an equation like

• \(w^2 - 15w + 54 = 0\).

To factor this, you're essentially asking: "What two numbers multiply to 54 but add up to -15?" The answer here is -9 and -6. This means the equation factors into:

• \((w - 9)(w - 6) = 0\).

Factoring effectively breaks down the equation into simpler parts, making it easier to find the values where the equation holds true. Thus, once factored, you can set each factor equal to zero, enabling you to find the solutions for \(w\).

Distributive Property

The distributive property is a fundamental building block in algebra used to simplify expressions and equations. It states that if you have an equation in the form \(a(b + c)\), it is equivalent to \(ab + ac\).
In context of the original equation \(54 = w(15-w)\), applying the distributive property allows us to expand the expression on the right side:

• First, multiply \(w\) by each term inside the parentheses:
  • \(w imes 15 = 15w\)
  • \(w imes (-w) = -w^2\)

So, we transform the equation to \(54 = 15w - w^2\). This step is crucial for setting the equation in a familiar quadratic form, which we can easily solve further.

Solving Equations

Solving equations, particularly quadratic ones, requires systematic strategies to find the variable's values that make the equation true. For the equation \(w^2 - 15w + 54 = 0\), the process involves a few manageable steps:

• Firstly, factor the quadratic, turning the equation into \((w - 9)(w - 6) = 0\).
• Next, apply the zero product property, which tells you if the product of two numbers is zero, at least one of them must be zero.

This property allows us to make two separate equations:

• \(w - 9 = 0\) → \(w = 9\)
• \(w - 6 = 0\) → \(w = 6\)

The solutions, \(w = 9\) and \(w = 6\), show where the initial quadratic equation's values hold true. By breaking down the problem, solving becomes a straightforward task.