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Magazine Chapter 1: Problem 25
Solve the inequality and express the solution in terms of intervals whenever possible. $$ x(2 x+3) \geq 5 $$
Short Answer
Expert verified
The solution is \([-\infty, -2.5]\cup [1, \infty)\).
Step by step solution
01
Expand the inequality
Start with the given inequality: \( x(2x + 3) \geq 5 \). Multiply \( x \) through the bracket to expand it: \( 2x^2 + 3x \geq 5 \).
02
Rearrange to form a quadratic inequality
Bring all terms to one side of the inequality to set it to zero: \( 2x^2 + 3x - 5 \geq 0 \). This forms a quadratic inequality in standard form.
03
Solve the related quadratic equation
Solve the equation \( 2x^2 + 3x - 5 = 0 \) by using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 3 \), and \( c = -5 \). Calculate the discriminant first: \( b^2 - 4ac = 3^2 - 4 \times 2 \times (-5) = 9 + 40 = 49 \). The roots are \( x = \frac{-3 \pm \sqrt{49}}{4} \).
04
Calculate the roots
Substitute and calculate the roots: \( x = \frac{-3 + 7}{4} = 1 \) and \( x = \frac{-3 - 7}{4} = -2.5 \). These roots are \( x = 1 \) and \( x = -2.5 \).
05
Determine the critical points and intervals
The critical points from the quadratic are \( x = -2.5 \) and \( x = 1 \). These divide the number line into three intervals: \([-\infty, -2.5)\), \([-2.5, 1] \), and \((1, \infty)\).
06
Test the intervals
Choose a test point from each interval and substitute it into \( 2x^2 + 3x - 5 \) to see if the result is \( \geq 0 \).- For \( x = -3 \) in \( [-\infty, -2.5) \), substituting gives \( 2(-3)^2 + 3(-3) - 5 = 18 - 9 - 5 = 4 \), which is \( \geq 0 \).- For \( x = 0 \) in \( [-2.5, 1] \), substituting gives \( 2(0)^2 + 3(0) - 5 = -5 \), which is \( < 0 \).- For \( x = 2 \) in \( (1, \infty) \), substituting gives \( 2(2)^2 + 3(2) - 5 = 8 + 6 - 5 = 9 \), which is \( \geq 0 \).
07
Write the solution interval
Based on the test points, the inequality holds for the intervals \( [-\infty, -2.5] \) and \([1, \infty) \), as these make the inequality true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in the form \( ax^2 + bx + c = 0 \). The "quadratic" term comes from "quad," meaning "square," as the highest power of the variable is 2.
For example, in your exercise, the quadratic equation we derived is \( 2x^2 + 3x - 5 = 0 \).
Here, \( a = 2 \), \( b = 3 \), and \( c = -5 \).
Quadratic equations are fundamental for modeling situations in physics, economics, and various fields where parabolic relationships occur.
For example, in your exercise, the quadratic equation we derived is \( 2x^2 + 3x - 5 = 0 \).
Here, \( a = 2 \), \( b = 3 \), and \( c = -5 \).
Quadratic equations are fundamental for modeling situations in physics, economics, and various fields where parabolic relationships occur.
- Standard Form: The equation is usually written as \( ax^2 + bx + c = 0 \).
- Solving Methods: Methods include factoring, completing the square, using graphs, or applying the quadratic formula.
Interval Notation
Interval notation is a method of representing solutions to inequalities graphically on a number line. Rather than representing solutions as a list of numbers, you use intervals.
In the exercise, after solving the quadratic inequality, the solutions were expressed as intervals \([-\infty, -2.5] \) and \([1, \infty) \).
This describes continuous sets of numbers rather than discrete points.
In the exercise, after solving the quadratic inequality, the solutions were expressed as intervals \([-\infty, -2.5] \) and \([1, \infty) \).
This describes continuous sets of numbers rather than discrete points.
- You use brackets \([ \) or \( ] \) to include an endpoint in the interval, indicating that the number itself is part of the set.
- Parentheses \(( \) or \() \) are used to indicate that the endpoint is not included in the interval.
- Infinity \( \infty \) and negative infinity \( -\infty \) always use parentheses because they are not actual numbers.
Discriminant
The discriminant is a key part of the quadratic formula, offering insights into the nature of the roots of a quadratic equation. It is calculated as \( b^2 - 4ac \).
In your problem, the discriminant for \( 2x^2 + 3x - 5 \) is calculated as follows: \[ b^2 - 4ac = 3^2 - 4 \times 2 \times (-5) = 9 + 40 = 49 \]A few important points about the discriminant are:
In your problem, the discriminant for \( 2x^2 + 3x - 5 \) is calculated as follows: \[ b^2 - 4ac = 3^2 - 4 \times 2 \times (-5) = 9 + 40 = 49 \]A few important points about the discriminant are:
- If the discriminant is positive, there are two distinct real roots.
- If it is zero, there is one real root (also known as a repeated or double root).
- If negative, the roots are complex and not real.
Inequality Solution
Solving a quadratic inequality involves finding where the quadratic expression is positive or negative. In your exercise, you were tasked with solving \( 2x^2 + 3x - 5 \geq 0 \).
Here's how it breaks down:
Here's how it breaks down:
- Find the Critical Points: Solve the quadratic \( 2x^2 + 3x - 5 = 0 \) using the quadratic formula to get critical points, where the expression changes sign: \( x = -2.5 \) and \( x = 1 \).
- Test Intervals: Check test points in the intervals divided by critical points to determine whether the interval satisfies the inequality.
- Formulate the Solution: Use these test results to write down the intervals where the inequality holds true, resulting in \([-\infty, -2.5] \) and \([1, \infty) \).
