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Chapter 3: Problem 57

Solve each absolute value equation. $$\left|x^{2}-x-6\right|=6$$

Short Answer

Expert verified
x = 4, x = -3, x = 0, x = 1

Step by step solution

01

- Understand the Absolute Value Equation

An absolute value equation \(\right|A\right|=B\) means that the expression inside the absolute value can be equal to \(B\) or \(-B\). Here, \(\right|x^{2}-x-6\right|=6\) implies \(x^2 - x - 6 = 6\) or \(x^2 - x - 6 = -6\).
02

- Set Up and Solve the First Equation

Solve \(x^2 - x - 6 = 6\): \[ x^2 - x - 6 = 6 \] \[ x^2 - x - 12 = 0 \] Factorize: \[ (x - 4)(x + 3) = 0 \] Thus, \[ x = 4 \ or \ x = -3 \].
03

- Set Up and Solve the Second Equation

Solve \(x^2 - x - 6 = -6\): \[ x^2 - x - 6 = -6 \] \[ x^2 - x = 0 \] Factorize: \[ x(x - 1) = 0 \] Thus, \[ x = 0 \ or \ x = 1 \].
04

- List All Solutions

Combine all the solutions from both equations: \[ x = 4, x = -3, x = 0, x = 1 \]

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Key Concepts

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

solving absolute value equations
Absolute value equations might seem tricky, but they can be simplified with a systematic approach. An absolute value equation \(\left|A\right|=B\) indicates that what's inside the absolute value (A) could be equal to B or -B. Understanding this concept is crucial as it lays the foundation for solving the equation.
For instance, consider the equation \(\left|x^{2}-x-6\right|=6\). This implies two separate equations: \(x^2 - x - 6 = 6\) and \(x^2 - x - 6 = -6\).
This understanding allows us to methodically break down the problem and solve it using standard algebraic methods, such as equations and factoring.
factoring quadratic equations
Factoring quadratic equations is a key skill in algebra, especially when solving absolute value equations. When given \(x^2 - x - 6 = 6\), we first rearrange the equation to standard form: \(x^2 - x - 12 = 0\). To factor, we look for two numbers that multiply to -12 and add to -1, which are -4 and 3. Thus, we factorize as so: \((x - 4)(x + 3) = 0\).
This results in two solutions: \(x = 4\) and \(x = -3\).
Dealing with the second equation, \(x^2 - x - 6 = -6\), we simplify it to \(x^2 - x = 0\), factoring it as \(x(x - 1) = 0\), yielding \(x = 0\) and \(x = 1\).
Mastering factoring is crucial for solving such algebraic problems with ease.
algebraic solutions
Combining algebraic solutions is a final step after solving individual equations. For our absolute value equation \(\left|x^{2}-x-6\right|=6\), we combined results from both sets of equations.
From \(x^2 - x - 12 = 0\), we got \(x = 4\) and \(x = -3\).
From \(x^2 - x = 0\), the solutions were \(x = 0\) and \(x = 1\).
Together, these yield \(x = 4, x = -3, x = 0, \) and \(x = 1\).
This step emphasizes the importance of understanding and applying algebraic methods uniformly to achieve a comprehensive set of solutions.

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Most popular questions from this chapter

Use division to write each rational expression in the form quotient \(+\) remainder/divisor. Use synthetic division when possible. $$\frac{x-1}{x+3}$$

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