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Magazine Chapter 1: Problem 60
Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\left(x^{2}-x\right)\left(x^{2}-5 x+6\right)<0$$
Short Answer
Expert verified
The solution is: \((-\infty, 0) \cup (0, 1) \cup (2, 3)\).
Step by step solution
01
Factor the terms
First, we factor each quadratic expression. The expression \(x^2 - x\) can be factored as \(x(x - 1)\). The quadratic \(x^2 - 5x + 6\) can be factored as \((x - 2)(x - 3)\). This makes the inequality:\[(x)(x - 1)(x - 2)(x - 3) < 0\]
02
Identify critical points
We identify the points where any factor might change sign, these are called critical points. Here, the critical points are solutions to each expression equal to zero: \(x = 0, 1, 2,\) and \(3\).
03
Determine sign changes between critical points
These critical points divide the number line into intervals: \((-fty, 0)\), \((0, 1)\), \((1, 2)\), \((2, 3)\), and \((3, fty)\). We now check the sign of the product \((x)(x - 1)(x - 2)(x - 3)\) in each interval by choosing a test point from each interval. For instance, for the interval \((-fty, 0)\), test \(x = -1\), and so forth.
04
Calculate the signs of each interval
Calculate the signs:- \((-fty, 0)\) : Choose \(x = -1\), signs are negative overall (\< 0).- \((0, 1)\) : Choose \(x = 0.5\), signs are negative overall (\< 0).- \((1, 2)\) : Choose \(x = 1.5\), signs are positive (\> 0).- \((2, 3)\) : Choose \(x = 2.5\), signs are negative overall (\< 0).- \((3, fty)\) : Choose \(x = 4\), signs are positive (\> 0).
05
Combine intervals with negative products
Based on the calculated signs, the inequality \((x)(x - 1)(x - 2)(x - 3) < 0\) is satisfied on intervals where the product is negative: \((-\infty, 0)\), \((0, 1)\), and \((2, 3)\).
06
Write the solution in interval notation
The solution to the inequality \(x^2 - x)(x^2 - 5x + 6) < 0\) is expressed in interval notation as: \((-\infty, 0) \cup (0, 1) \cup (2, 3)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
Factoring quadratics is a key step in solving quadratic inequalities. Just like breaking down a complex machine into simpler parts, factoring helps you see each component clearly. The process involves expressing the quadratic polynomial as a product of its factors. For example, with the expressions provided:
- The quadratic expression \(x^2 - x\) factors into \(x(x - 1)\). This means finding numbers that, when multiplied, give you the original expression and add up to the coefficient of the middle term (if present).
- Similarly, \(x^2 - 5x + 6\) factors into \((x - 2)(x - 3)\). These factors come from finding numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of \(x\)).
Critical Points
Critical points in the context of inequalities are values where a polynomial changes its sign. These points occur where the expression equals zero. Once a quadratic has been factored, finding its critical points is straightforward. Simply set each factor equal to zero:
- For \(x(x - 1)(x - 2)(x - 3)\), solve each factor:
- \(x = 0\)
- \(x - 1 = 0\), so \(x = 1\)
- \(x - 2 = 0\), so \(x = 2\)
- \(x - 3 = 0\), so \(x = 3\)
Interval Notation
Interval notation is like a shorthand method for describing sets of numbers. It provides a convenient way of representing the intervals that satisfy an inequality. In this method:
- An open interval, such as \((a, b)\), includes all numbers greater than \(a\) and less than \(b\), but not \(a\) or \(b\) themselves.
- A closed interval, written as \([a, b]\), includes \(a\), \(b\), and all the numbers in between.
- If an endpoint extends to infinity, like \((-fty, a)\) or \((b, fty)\), it means the interval extends endlessly in the specified direction.
Sign Analysis
Sign analysis is a critical part of solving inequalities. The goal is to determine whether the product expression \((x)(x - 1)(x - 2)(x - 3)\) is positive or negative in each identified interval. You can achieve this by choosing test points within each interval and plugging them into the factored expression:
- For \((-fty, 0)\), pick a point like \(x = -1\). Calculate the sign of the expression, and you'll find it negative.
- Continue with other intervals: \((0, 1)\), choose \(x = 0.5\); \((1, 2)\), choose \(x = 1.5\); \((2, 3)\), choose \(x = 2.5\); and \((3, fty)\), choose \(x = 4\).
