Five Times Of A Positive Integer Is Less Than Twice Its Square By 3. Find The Integer.​

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Introduction

In this article, we will delve into the world of algebra and solve a problem that involves a positive integer. The problem states that five times of a positive integer is less than twice its square by 3. We will use algebraic equations to represent this problem and solve for the value of the positive integer.

Understanding the Problem

Let's break down the problem statement:

  • Five times of a positive integer: 5x
  • Twice its square: 2x^2
  • Less than twice its square by 3: 2x^2 - 3

The problem statement can be represented as an equation:

5x < 2x^2 - 3

Representing the Problem as an Equation

To solve this problem, we need to represent it as an equation. Let's rewrite the problem statement as an equation:

5x < 2x^2 - 3

We can rewrite this equation as:

2x^2 - 5x - 3 > 0

Solving the Quadratic Equation

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 2, b = -5, and c = -3. Plugging these values into the quadratic formula, we get:

x = (5 ± √((-5)^2 - 4(2)(-3))) / (2(2)) x = (5 ± √(25 + 24)) / 4 x = (5 ± √49) / 4 x = (5 ± 7) / 4

Finding the Solutions

Now that we have the solutions to the quadratic equation, we need to find the values of x that satisfy the original inequality. We can do this by plugging in the values of x into the original inequality:

5x < 2x^2 - 3

We can try plugging in x = (5 + 7) / 4 and x = (5 - 7) / 4 into the original inequality to see which one satisfies the inequality.

Checking the Solutions

Let's try plugging in x = (5 + 7) / 4 into the original inequality:

5((5 + 7) / 4) < 2((5 + 7) / 4)^2 - 3 5(12 / 4) < 2(12 / 4)^2 - 3 15 < 2(36 / 16) - 3 15 < 9/4 - 3 15 < -3/4 15 > -3/4

This solution does not satisfy the original inequality.

Now let's try plugging in x = (5 - 7) / 4 into the original inequality:

5((5 - 7) / 4) < 2((5 - 7) / 4)^2 - 3 5(-2 / 4) < 2(-2 / 4)^2 - 3 -5/2 < 2(1/16) - 3 -5/2 < 1/8 - 3 -5/2 < -23/8 -5/2 > -23/8

This solution also does not satisfy the original inequality.

However, we can try plugging in x = 4 into the original inequality:

5(4) < 2(4)^2 - 3 20 < 32 - 3 20 < 29 20 < 29

This solution satisfies the original inequality.

Conclusion

In this article, we solved a problem that involved a positive integer. We represented the problem as an equation, solved the quadratic equation, and found the solutions to the equation. We then checked the solutions to see which one satisfied the original inequality. We found that x = 4 satisfies the original inequality.

Final Answer

The final answer is x = 4.

Key Takeaways

  • We can represent a problem as an equation and solve it using algebraic equations.
  • We can use the quadratic formula to solve quadratic equations.
  • We need to check the solutions to an equation to see which one satisfies the original inequality.

Common Core State Standards

  • 8.EE.8: Analyze and solve linear equations and pairs of linear equations in one variable.
  • 8.EE.7: Solve a simple system consisting of a linear equation and a linear inequality in one variable.
  • 8.EE.6: Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to represent the relationship between the quantities.

CBSE Board X

  • This problem is relevant to the CBSE Board X curriculum, which covers algebra and equations.
  • Students in the CBSE Board X curriculum should be able to represent a problem as an equation, solve the equation, and check the solutions to see which one satisfies the original inequality.
    Q&A: Solving the Mystery of the Positive Integer =====================================================

Introduction

In our previous article, we solved a problem that involved a positive integer. We represented the problem as an equation, solved the quadratic equation, and found the solutions to the equation. We then checked the solutions to see which one satisfied the original inequality. In this article, we will answer some frequently asked questions about the problem.

Q: What is the problem statement?

A: The problem statement is: Five times of a positive integer is less than twice its square by 3.

Q: How do we represent the problem as an equation?

A: We can represent the problem as an equation by writing:

5x < 2x^2 - 3

Q: How do we solve the quadratic equation?

A: We can solve the quadratic equation by using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 2, b = -5, and c = -3.

Q: What are the solutions to the quadratic equation?

A: The solutions to the quadratic equation are:

x = (5 ± √49) / 4 x = (5 ± 7) / 4

Q: How do we check the solutions to see which one satisfies the original inequality?

A: We can check the solutions by plugging them into the original inequality:

5x < 2x^2 - 3

We can try plugging in x = (5 + 7) / 4 and x = (5 - 7) / 4 into the original inequality to see which one satisfies the inequality.

Q: What is the final answer?

A: The final answer is x = 4.

Q: What are some key takeaways from this problem?

A: Some key takeaways from this problem are:

  • We can represent a problem as an equation and solve it using algebraic equations.
  • We can use the quadratic formula to solve quadratic equations.
  • We need to check the solutions to an equation to see which one satisfies the original inequality.

Q: How does this problem relate to the CBSE Board X curriculum?

A: This problem is relevant to the CBSE Board X curriculum, which covers algebra and equations. Students in the CBSE Board X curriculum should be able to represent a problem as an equation, solve the equation, and check the solutions to see which one satisfies the original inequality.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem are:

  • Not representing the problem as an equation.
  • Not solving the quadratic equation correctly.
  • Not checking the solutions to see which one satisfies the original inequality.

Q: How can I practice solving problems like this?

A: You can practice solving problems like this by:

  • Working on similar problems in your textbook or online resources.
  • Creating your own problems and solving them.
  • Joining a study group or seeking help from a tutor.

Conclusion

In this article, we answered some frequently asked questions about the problem of solving a positive integer. We hope that this article has been helpful in clarifying any confusion and providing additional support for students who are working on this problem.