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Chapter 3: Problem 55

Solve each absolute value equation. $$\left|v^{2}-3 v\right|=5 v$$

Short Answer

Expert verified
The solutions are \( v = 0 \), \( v = 8 \), and \( v = -2 \).

Step by step solution

01

Understand the Absolute Value Equation

An absolute value equation \(|x| = a\) translates to two linear equations: \( x = a \) and \( x = -a \). Hence, rewrite \(|v^{2} - 3v| = 5v\) as \( v^{2} - 3v = 5v \) and \( v^{2} - 3v = -5v \).
02

Solve the First Equation

Solve \( v^{2} - 3v = 5v \). 1. Subtract \(5v\) from both sides to get: \( v^{2} - 8v = 0 \). 2. Factor out \(v \): \( v(v - 8) = 0 \). 3. Set each factor equal to zero: \( v = 0 \) or \( v - 8 = 0 \). 4. Solve for \(v\): \( v = 0 \) or \( v = 8 \).
03

Solve the Second Equation

Solve \( v^{2} - 3v = -5v \). 1. Add \(5v\) to both sides to get: \( v^{2} + 2v = 0 \). 2. Factor out \(v \): \( v(v + 2) = 0 \). 3. Set each factor equal to zero: \( v = 0 \) or \( v + 2 = 0 \). 4. Solve for \(v\): \( v = 0 \) or \( v = -2 \).
04

Combine Solutions and Verify

Combine all solutions from both equations: \( v = 0 \), \( v = 8 \), and \( v = -2 \). Verify each solution by substituting back into the original equation \(|v^{2} - 3v| = 5v\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving absolute value equations
Solving absolute value equations involves understanding the nature of absolute values. Absolute value, denoted by vertical bars \(|x| \), represents the distance of a number from zero without considering direction. For any absolute value equation \(|x| = a\), where \(a > 0\), there are always two possible values for \[x\]: \(x = a\) and \(x = -a\).
Given the equation \(|v^{2} - 3v| = 5v\), we translate it into two separate linear equations:
  • \ v^{2} - 3v = 5v
  • \ v^{2} - 3v = -5v
This transformation helps simplify and solve these equations separately to find all possible solutions.
factoring quadratic equations
Factoring quadratic equations is a key step in solving equations that include squares of variables \((v^2)\). To factor a quadratic equation means to rewrite it as a product of its factors. For example, in the first equation \ v^{2} - 3v = 5v \, we first rearrange it to \(v^{2} - 8v = 0\). Factoring out the common variable \(v\), we get:
  • \

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