Solve For The Value Of W.(w-8)(2W-1)

Solve For The Value Of W.(w-8)(2W-1)

When encountering algebraic expressions like W.(w-8)(2W-1), understanding how to simplify and solve for the unknown variable \( W \) is essential. Whether you're a student working through algebra homework or a professional applying mathematical concepts, mastering the process of solving such expressions enhances your problem-solving skills. This article provides a comprehensive guide on how to approach, simplify, and solve for \( W \) in the expression W.(w-8)(2W-1), with detailed explanations, step-by-step procedures, and tips for efficient problem-solving.

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Understanding the Expression: W.(w-8)(2W-1)

Before diving into solving, it's important to understand the structure of the expression:

• The expression is multiplicative, involving the variable \( W \) and two binomials \((w-8)\) and \((2W-1)\).


• The notation suggests that \( W \) and \( w \) may be variables, but for the purpose of solving for \( W \), we'll treat \( w \) as a known value or parameter unless specified otherwise.


• The goal is to find the value(s) of \( W \) that satisfy an equation involving this expression.

Interpreting the Problem: Setting the Expression Equal to a Value

In algebra, an expression like W.(w-8)(2W-1) is often set equal to a value or another expression to form an equation. For example:

```plaintext
W.(w-8)(2W-1) = K
```

where \( K \) is a known constant, or zero in case of an equation to find roots.

Common problem types include:

• Solving for \( W \) when the expression equals a constant


• Simplifying the expression for further analysis


• Finding the roots or solutions that satisfy the equation

```plaintext
W.(w-8)(2W-1) = 0
```

which is a typical starting point for solving such expressions.

---

Step-by-Step Guide to Solve W.(w-8)(2W-1) = 0

When an expression is set equal to zero, the Zero Product Property applies. This property states that if:

```plaintext
A B C = 0
```

then at least one of the factors must be zero:

```plaintext
A = 0 OR B = 0 OR C = 0
```

Applying this to the expression:

```plaintext
W.(w-8)(2W-1) = 0
```

the solutions are obtained by setting each factor equal to zero:

1. \( W = 0 \)


2. \( w - 8 = 0 \)


3. \( 2W - 1 = 0 \)

This is straightforward; one solution is:

```plaintext
W = 0
```

Solution 2: \( w - 8 = 0 \)

Since \( w \) appears to be a parameter, this solution depends on knowing the value of \( w \). If \( w \) is known, then:

```plaintext
w = 8
```

If \( w \) is unknown, this equation indicates that for the original expression to be zero, \( w \) must be 8.

Solution 3: \( 2W - 1 = 0 \)

Solve for \( W \):

```plaintext
2W = 1
W = \frac{1}{2}
```

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Summary of Solutions

| Solution | Description | Notes |
|---|---|---|
| \( W = 0 \) | Direct solution from the first factor | Always valid regardless of \( w \) |
| \( w = 8 \) | Conditions on \( w \) for the expression to be zero | Depends on \( w \) being known or specified |
| \( W = \frac{1}{2} \) | From solving \( 2W - 1 = 0 \) | Valid for any \( w \) unless constrained |

---

Extending Beyond Zero: Solving General Equations

Often, you may encounter equations where the expression equals a non-zero value:

```plaintext
W.(w-8)(2W-1) = K
```

To solve for \( W \), follow these steps:

Step 1: Expand or Rearrange the Equation

Depending on the value of \( K \), you might need to expand the expression:

```plaintext
W (w - 8) (2W - 1) = K
```

Step 2: Expand the Expression

First, expand \( (w - 8)(2W - 1) \):

```plaintext
(w - 8)(2W - 1) = w2W - w1 - 82W + 81 = 2wW - w - 16W + 8
```

Then, multiply by \( W \):

```plaintext
W (2wW - w - 16W + 8) = 2wW^2 - wW - 16W^2 + 8W
```

Step 3: Form a Quadratic Equation

Bring all terms to one side:

```plaintext
2wW^2 - wW - 16W^2 + 8W - K = 0
```

Group similar terms:

```plaintext
(2wW^2 - 16W^2) + (-wW + 8W) - K = 0
```

Factor where possible:

```plaintext
2W^2(w - 8) + W(-w + 8) - K = 0
```

This is a quadratic in \( W \):

```plaintext
A W^2 + B W + C = 0
```

where:

• \( A = 2(w - 8) \)


• \( B = -w + 8 \)


• \( C = -K \)

The solutions for \( W \):

```plaintext
W = \frac{-B \pm \sqrt{B^2 - 4 A C}}{2A}
```

Plug in \( A, B, C \):

```plaintext
W = \frac{-(-w + 8) \pm \sqrt{(-w + 8)^2 - 4 2(w - 8) (-K)}}{2 2(w - 8)}
```

Simplify numerator and denominator accordingly.

---

Practical Tips for Solving Algebraic Expressions

• Always check for extraneous solutions: When solving polynomial equations, verify solutions in the original equation.
• Factor first: Simplify the expression by factoring before applying the quadratic formula.
• Use substitution: For complex expressions, substitution can make solving easier.
• Be aware of domain restrictions: For example, division by zero is undefined; check if any solutions violate such restrictions.
• Practice with different values: To build intuition, solve similar problems with specific values of \( w \) and \( K \).

Sample Problems and Solutions

Example 1: Solve \( W.(w-8)(2W-1) = 0 \) with \( w = 10 \)

Solution:

• Factor set to zero:

1. \( W = 0 \)


2. \( 10 - 8 = 2 \neq 0 \) (so no solution from this factor unless \( w = 8 \))


3. \( 2W - 1 = 0 \Rightarrow W = \frac{1}{2} \)

```plaintext
W = 0 \quad \text{or} \quad W = \frac{1}{2}
```

Example 2: Solve \( W.(w-8)(2W-1) = 5 \) with \( w = 12 \)

Solution:

• Expand:

• Express:

• Simplify:

• Expand:

• Rearrange:

• Use quadratic formula:

\( W = \frac{4 \pm \sqrt{16 + 160}}{16} \)

\( W = \frac{4 \pm \sqrt{176}}{16} \)

\( W = \frac{4 \pm 13.266}{16} \)

• Solutions:

\( W = \frac{4 - 13