Solve The Polynomial Inequality And Graph The Solution Set On A Real Number Line. Express The Solution Set In Interval Notation.$\[ 5x^2 + 32x - 21 \ \textless \ 0 \\]Solve The Inequality. What Is The Solution Set? Select The Correct Choice

by ADMIN 243 views

Introduction

Polynomial inequalities are a fundamental concept in algebra, and solving them requires a deep understanding of the properties of polynomials and their behavior on the real number line. In this article, we will focus on solving the polynomial inequality 5x2+32xβˆ’21<0{5x^2 + 32x - 21 < 0} and graphing the solution set on a real number line. We will also express the solution set in interval notation.

Understanding the Problem

The given inequality is a quadratic inequality, which means it involves a quadratic expression on the left-hand side. The quadratic expression is 5x2+32xβˆ’21{5x^2 + 32x - 21}, and we want to find the values of x{x} for which this expression is less than zero.

Step 1: Factor the Quadratic Expression

To solve the inequality, we first need to factor the quadratic expression. We can do this by finding two numbers whose product is βˆ’105{-105} (the product of the coefficient of x2{x^2} and the constant term) and whose sum is 32{32} (the coefficient of x{x}). These numbers are 35{35} and βˆ’3{-3}, so we can write the quadratic expression as:

5x2+32xβˆ’21=(5xβˆ’7)(x+3){5x^2 + 32x - 21 = (5x - 7)(x + 3)}

Step 2: Find the Critical Points

The critical points are the values of x{x} that make the quadratic expression equal to zero. In this case, we have two critical points: x=75{x = \frac{7}{5}} and x=βˆ’3{x = -3}.

Step 3: Test the Intervals

To determine the solution set, we need to test the intervals between the critical points. We can do this by choosing a test point in each interval and plugging it into the quadratic expression.

Interval 1: x<βˆ’3{x < -3}

Let's choose a test point, say x=βˆ’4{x = -4}. Plugging this value into the quadratic expression, we get:

(5(βˆ’4)βˆ’7)(βˆ’4+3)=(βˆ’25βˆ’7)(βˆ’1)=βˆ’32β‹…βˆ’1=32>0{(5(-4) - 7)(-4 + 3) = (-25 - 7)(-1) = -32 \cdot -1 = 32 > 0}

Since the quadratic expression is positive in this interval, we can conclude that the solution set does not include any values of x{x} less than βˆ’3{-3}.

Interval 2: βˆ’3<x<75{-3 < x < \frac{7}{5}}

Let's choose a test point, say x=0{x = 0}. Plugging this value into the quadratic expression, we get:

(5(0)βˆ’7)(0+3)=(βˆ’7)(3)=βˆ’21<0{(5(0) - 7)(0 + 3) = (-7)(3) = -21 < 0}

Since the quadratic expression is negative in this interval, we can conclude that the solution set includes all values of x{x} between βˆ’3{-3} and 75{\frac{7}{5}}.

Interval 3: x>75{x > \frac{7}{5}}

Let's choose a test point, say x=2{x = 2}. Plugging this value into the quadratic expression, we get:

(5(2)βˆ’7)(2+3)=(3)(5)=15>0{(5(2) - 7)(2 + 3) = (3)(5) = 15 > 0}

Since the quadratic expression is positive in this interval, we can conclude that the solution set does not include any values of x{x} greater than 75{\frac{7}{5}}.

Conclusion

Based on our analysis, we can conclude that the solution set to the inequality 5x2+32xβˆ’21<0{5x^2 + 32x - 21 < 0} is the interval (βˆ’3,75){(-3, \frac{7}{5})}. This means that the values of x{x} that satisfy the inequality are all real numbers between βˆ’3{-3} and 75{\frac{7}{5}}, but not including βˆ’3{-3} and 75{\frac{7}{5}}.

Graphing the Solution Set

To graph the solution set, we can use a number line and mark the critical points βˆ’3{-3} and 75{\frac{7}{5}}. We can then test the intervals between these points by choosing test points and plugging them into the quadratic expression. The solution set will be the interval (βˆ’3,75){(-3, \frac{7}{5})}, which can be represented graphically as:

Interval Notation

The solution set can be expressed in interval notation as (βˆ’3,75){(-3, \frac{7}{5})}. This notation indicates that the solution set includes all real numbers between βˆ’3{-3} and 75{\frac{7}{5}}, but not including βˆ’3{-3} and 75{\frac{7}{5}}.

Final Answer

Introduction

In our previous article, we explored the concept of solving polynomial inequalities and graphing the solution set on a real number line. We also expressed the solution set in interval notation. In this article, we will provide a Q&A guide to help you better understand the concept and apply it to different problems.

Q: What is a polynomial inequality?

A: A polynomial inequality is an inequality that involves a polynomial expression on the left-hand side and a constant or variable on the right-hand side. For example, 5x2+32xβˆ’21<0{5x^2 + 32x - 21 < 0} is a polynomial inequality.

Q: How do I solve a polynomial inequality?

A: To solve a polynomial inequality, you need to follow these steps:

  1. Factor the quadratic expression, if possible.
  2. Find the critical points by setting the quadratic expression equal to zero.
  3. Test the intervals between the critical points by choosing test points and plugging them into the quadratic expression.
  4. Determine the solution set based on the results of the interval testing.

Q: What are critical points?

A: Critical points are the values of x{x} that make the quadratic expression equal to zero. They are the points where the graph of the quadratic expression intersects the x-axis.

Q: How do I test the intervals?

A: To test the intervals, you need to choose a test point in each interval and plug it into the quadratic expression. If the result is positive, the interval is not part of the solution set. If the result is negative, the interval is part of the solution set.

Q: What is interval notation?

A: Interval notation is a way of expressing the solution set in a compact and concise form. It uses parentheses or brackets to indicate the endpoints of the interval.

Q: How do I graph the solution set?

A: To graph the solution set, you need to use a number line and mark the critical points. You can then test the intervals between these points by choosing test points and plugging them into the quadratic expression. The solution set will be the interval that satisfies the inequality.

Q: What are some common mistakes to avoid when solving polynomial inequalities?

A: Some common mistakes to avoid when solving polynomial inequalities include:

  • Not factoring the quadratic expression correctly
  • Not finding all the critical points
  • Not testing all the intervals
  • Not expressing the solution set in interval notation

Q: How can I apply polynomial inequalities to real-world problems?

A: Polynomial inequalities can be applied to a wide range of real-world problems, including:

  • Modeling population growth and decline
  • Analyzing the behavior of physical systems
  • Optimizing functions and minimizing costs
  • Solving problems in engineering, economics, and other fields

Conclusion

Solving polynomial inequalities requires a deep understanding of the properties of polynomials and their behavior on the real number line. By following the steps outlined in this article and avoiding common mistakes, you can master the concept and apply it to a wide range of problems.

Final Tips

  • Practice, practice, practice! Solving polynomial inequalities requires a lot of practice to become proficient.
  • Use technology, such as graphing calculators or computer software, to help you visualize the solution set and check your work.
  • Read and understand the problem carefully before starting to solve it.
  • Check your work carefully to avoid making mistakes.

Common Polynomial Inequalities

Here are some common polynomial inequalities that you may encounter:

  • x2+4x+4<0{x^2 + 4x + 4 < 0}
  • x2βˆ’6x+8>0{x^2 - 6x + 8 > 0}
  • 2x2+3xβˆ’1<0{2x^2 + 3x - 1 < 0}
  • x2βˆ’2xβˆ’3>0{x^2 - 2x - 3 > 0}

Solutions to Common Polynomial Inequalities

Here are the solutions to the common polynomial inequalities listed above:

  • x2+4x+4<0{x^2 + 4x + 4 < 0} has no solution
  • x2βˆ’6x+8>0{x^2 - 6x + 8 > 0} has solution (2,4){(2, 4)}
  • 2x2+3xβˆ’1<0{2x^2 + 3x - 1 < 0} has solution (βˆ’1,1){(-1, 1)}
  • x2βˆ’2xβˆ’3>0{x^2 - 2x - 3 > 0} has solution (βˆ’1,3){(-1, 3)}

Final Answer

The final answer is: (βˆ’3,75)\boxed{(-3, \frac{7}{5})}