Two Positive Integers Have A Product Of 176. One Integer Is 5 Less Than The Other Integer. Which Equation Can Be Used To Find The Value Of X X X , The Greater Integer?A. X 2 + 5 = 176 X^2 + 5 = 176 X 2 + 5 = 176 B. X ( X + 5 ) = 176 X(x + 5) = 176 X ( X + 5 ) = 176 C. $x(x - 5) =

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Introduction

In mathematics, solving equations is a fundamental concept that helps us find the value of unknown variables. In this article, we will explore a problem that involves two positive integers with a product of 176, where one integer is 5 less than the other. Our goal is to find the equation that can be used to determine the value of xx, the greater integer.

Understanding the Problem

Let's break down the problem and understand what is being asked. We have two positive integers, and their product is 176. This means that if we multiply these two integers together, we get 176. Additionally, we know that one integer is 5 less than the other. This gives us a relationship between the two integers.

Defining the Integers

Let's define the two integers as xx and xβˆ’5x - 5. Since xx is the greater integer, we can represent the smaller integer as xβˆ’5x - 5. This is because the smaller integer is 5 less than the greater integer.

Formulating the Equation

Now that we have defined the integers, we can formulate the equation that represents their product. Since the product of the two integers is 176, we can write the equation as:

x(xβˆ’5)=176x(x - 5) = 176

This equation represents the product of the two integers, where xx is the greater integer and xβˆ’5x - 5 is the smaller integer.

Analyzing the Options

Let's analyze the options provided to determine which equation can be used to find the value of xx, the greater integer.

  • Option A: x2+5=176x^2 + 5 = 176
  • Option B: x(x+5)=176x(x + 5) = 176
  • Option C: x(xβˆ’5)=176x(x - 5) = 176

Option A

Option A is x2+5=176x^2 + 5 = 176. This equation is not correct because it does not represent the product of the two integers. Instead, it represents the square of the greater integer plus 5.

Option B

Option B is x(x+5)=176x(x + 5) = 176. This equation is also not correct because it represents the product of the greater integer and the sum of the greater integer and 5, not the product of the two integers.

Option C

Option C is x(xβˆ’5)=176x(x - 5) = 176. This equation is correct because it represents the product of the two integers, where xx is the greater integer and xβˆ’5x - 5 is the smaller integer.

Conclusion

In conclusion, the equation that can be used to find the value of xx, the greater integer, is x(xβˆ’5)=176x(x - 5) = 176. This equation represents the product of the two integers, where xx is the greater integer and xβˆ’5x - 5 is the smaller integer.

Solving the Equation

To solve the equation x(xβˆ’5)=176x(x - 5) = 176, we can start by expanding the left-hand side of the equation:

x2βˆ’5x=176x^2 - 5x = 176

Next, we can rearrange the equation to form a quadratic equation:

x2βˆ’5xβˆ’176=0x^2 - 5x - 176 = 0

We can solve this quadratic equation using the quadratic formula:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a=1a = 1, b=βˆ’5b = -5, and c=βˆ’176c = -176. Plugging these values into the quadratic formula, we get:

x=βˆ’(βˆ’5)Β±(βˆ’5)2βˆ’4(1)(βˆ’176)2(1)x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-176)}}{2(1)}

Simplifying the expression, we get:

x=5Β±25+7042x = \frac{5 \pm \sqrt{25 + 704}}{2}

x=5Β±7292x = \frac{5 \pm \sqrt{729}}{2}

x=5Β±272x = \frac{5 \pm 27}{2}

This gives us two possible solutions for xx:

x=5+272=16x = \frac{5 + 27}{2} = 16

x=5βˆ’272=βˆ’11x = \frac{5 - 27}{2} = -11

However, since we are looking for a positive integer, we can discard the negative solution and conclude that x=16x = 16.

Final Answer

Q: What is the problem asking for?

A: The problem is asking us to find the equation that can be used to determine the value of xx, the greater integer, given that two positive integers have a product of 176 and one integer is 5 less than the other.

Q: How do we define the integers?

A: We define the two integers as xx and xβˆ’5x - 5. Since xx is the greater integer, we can represent the smaller integer as xβˆ’5x - 5.

Q: What is the correct equation to use?

A: The correct equation to use is x(xβˆ’5)=176x(x - 5) = 176. This equation represents the product of the two integers, where xx is the greater integer and xβˆ’5x - 5 is the smaller integer.

Q: How do we solve the equation?

A: To solve the equation x(xβˆ’5)=176x(x - 5) = 176, we can start by expanding the left-hand side of the equation:

x2βˆ’5x=176x^2 - 5x = 176

Next, we can rearrange the equation to form a quadratic equation:

x2βˆ’5xβˆ’176=0x^2 - 5x - 176 = 0

We can solve this quadratic equation using the quadratic formula:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: What are the steps to solve the quadratic equation?

A: The steps to solve the quadratic equation are:

  1. Expand the left-hand side of the equation.
  2. Rearrange the equation to form a quadratic equation.
  3. Use the quadratic formula to solve the equation.

Q: What is the quadratic formula?

A: The quadratic formula is:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: What are the values of aa, bb, and cc in the quadratic formula?

A: In the quadratic formula, a=1a = 1, b=βˆ’5b = -5, and c=βˆ’176c = -176.

Q: How do we plug these values into the quadratic formula?

A: We plug the values of aa, bb, and cc into the quadratic formula as follows:

x=βˆ’(βˆ’5)Β±(βˆ’5)2βˆ’4(1)(βˆ’176)2(1)x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-176)}}{2(1)}

Q: What is the final solution to the equation?

A: The final solution to the equation is x=16x = 16.

Q: Why do we discard the negative solution?

A: We discard the negative solution because we are looking for a positive integer.

Q: What is the final answer to the problem?

A: The final answer to the problem is x=16x = 16.

Conclusion

In conclusion, solving for the greater integer involves defining the integers, formulating the equation, and solving the quadratic equation using the quadratic formula. By following these steps, we can find the value of xx, the greater integer.