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Chapter 3: Problem 58

Solve each absolute value equation. $$\left|2 x^{2}-x-2\right|=1$$

Short Answer

Expert verified
The solutions are \( x = 1, x = -\frac{3}{2}, x = -\frac{1}{2} \).

Step by step solution

01

- Understand Absolute Value Definition

Absolute value equation \(|A| = B\) means that \(A = B\) or \(A = -B\). Thus, for \(|2x^2 - x - 2| = 1\), the equations are \(2x^2 - x - 2 = 1\) and \(2x^2 - x - 2 = -1\).
02

- Solve the First Equation

Rearrange \(2x^2 - x - 2 = 1\) to \(2x^2 - x - 3 = 0\). Solve this quadratic equation using the quadratic formula \((x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})\), where \(a = 2, b = -1, c = -3\). This yields \(x = 1\) and \(x = -\frac{3}{2}\).
03

- Solve the Second Equation

Rearrange \(2x^2 - x - 2 = -1\) to \(2x^2 - x - 1 = 0\). Solve this quadratic equation using the quadratic formula where \(a = 2, b = -1, c = -1\). This yields \(x = 1\) and \(x = -\frac{1}{2}\).
04

- Combine Solutions

The solutions from both equations are \(x = 1\), \(x = -\frac{3}{2}\), and \(x = -\frac{1}{2}\). All of these solutions are valid.

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Key Concepts

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

absolute value
Absolute value refers to the distance of a number from zero on the number line. It is always a non-negative value. For example, the absolute value of both 3 and -3 is 3.
When solving an absolute value equation like \(|2x^2 - x - 2| = 1\), it means we need to find the points at which the expression inside the absolute value is equal to 1 or -1. Thus, \(|A| = B\) results in two equations: \(A = B\) and \(A = -B\).
This crucial step transforms the absolute value equation into two separate quadratic equations, making it easier to solve.
quadratic equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\) where \(a, b,\) and \(c\) are constants. These equations have a special property: they always result in a parabolic graph.
In the given problem, the absolute value equation splits into two quadratic equations:
  • \(2x^2 - x - 3 = 0\)
  • \(2x^2 - x - 1 = 0\)
These equations need to be solved separately to find the possible values of \(x\).
quadratic formula
The quadratic formula is used to find the solutions of a quadratic equation \(ax^2 + bx + c = 0\). It is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula comes handy especially when factoring is difficult or not possible.
For the first equation \(2x^2 - x - 3 = 0\), applying the quadratic formula with \(a = 2, b = -1, \text{and} c = -3\) gives:
\[x = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}\]
This results in two solutions: \(x = 1\) and \(x = -1.5\).
Similarly, for the second equation \(2x^2 - x - 1 = 0\), using \(a = 2, b = -1, \text{and} c = -1\) gives:
\[x = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}\]
Resulting in solutions \(x = 1\) and \(x = -0.5\).
roots of quadratic equations
The solutions to a quadratic equation are known as its roots. These roots can be real or complex, depending on the discriminant \((b^2 - 4ac)\). If the discriminant is positive, the quadratic equation has two distinct real roots.
In our case:
  • For \(2x^2 - x - 3 = 0\), the discriminant \(1 + 24\) is positive, yielding roots \(x = 1\) and \(x = -1.5\).
  • For \(2x^2 - x - 1 = 0\), the discriminant \(1 + 8\) is positive, giving roots \(x = 1\) and \(x = -0.5\).
Combining the roots from both quadratic equations, we get the valid solutions for the original problem: \(x = 1, x = -1.5, \text{and} x = -0.5\).

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