Problem 12 - Elementary and Intermediate Algebra

Question

CONCEPTS Find two integers whose A. product is 10 and whose sum is 7 . B. product is 8 and whose sum is \(-6\) C. product is \(-6\) and whose sum is 1 D. product is \(-9\) and whose sum is \(-8\).

Step-by-Step Solution

Answer

A: (5, 2) or (2, 5); B: (-4, -2) or (-2, -4); C: (3, -2) or (-2, 3); D: (-9, 1) or (1, -9).

Step 1: Understand the Problem

We need to find two integers (let's call them \(x\) and \(y\)) such that their product and sum satisfy the given conditions for each part of the exercise.

Step 2: Solve Part A

For part A, we need to find two integers whose product is 10 and sum is 7. Set up the equations: 1. \(xy = 10\)2. \(x + y = 7\)Solve the second equation for \(y\): \(y = 7 - x\).Substitute in the first equation:\(x(7-x) = 10\) which simplifies to \(x^2 - 7x + 10 = 0\).Factor the quadratic: \((x - 5)(x - 2) = 0\) Thus, \(x = 5\) or \(x = 2\). Checking gives pairs (5, 2) or (2, 5) as solutions.

Step 3: Solve Part B

For part B, the objective is to find two integers whose product is 8 and sum is \(-6\). Set up the equations: 1. \(xy = 8\)2. \(x + y = -6\)Solve the second equation for \(y\): \(y = -6 - x\).Substitute in the first equation:\(x(-6-x) = 8\), simplifies to \(x^2 + 6x + 8 = 0\).Factor the quadratic:\((x + 4)(x + 2) = 0\) Thus, \(x = -4\) or \(x = -2\). These give pairs (-4, -2) or (-2, -4) as solutions.

Step 4: Solve Part C

For part C, find two integers whose product is \(-6\) and sum is 1. Set up the equations:1. \(xy = -6\)2. \(x + y = 1\)Solve for \(y\):\(y = 1 - x\).Substitute in the first equation:\(x(1-x) = -6\), simplifies to \(x^2 - x - 6 = 0\).Factor the quadratic:\((x - 3)(x + 2) = 0\)Thus, \(x = 3\) or \(x = -2\). Solutions are (3, -2) or (-2, 3).

Step 5: Solve Part D

For part D, find two numbers whose product is \(-9\) and sum is \(-8\). Set up the equations:1. \(xy = -9\)2. \(x + y = -8\)Solve for \(y\):\(y = -8 - x\).Substitute in the first equation:\(x(-8-x) = -9\), simplifies to \(x^2 + 8x - 9 = 0\).Factor the quadratic:\((x + 9)(x - 1) = 0\)Thus, \(x = -9\) or \(x = 1\). Solutions are (-9, 1) or (1, -9).

Key Concepts

Integer Equations

Integer equations involve mathematical expressions that equate two or more integers, often represented as variables. In these types of problems, we are tasked with finding integer solutions that satisfy both the sum and product of two numbers. These numbers are often denoted as \(x\) and \(y\). In practice, this type of problem is frequently encountered in algebra courses, as it teaches students how to set up and manipulate equations to uncover solutions.

• Identify the relationships between the variables and their given properties, like sums and products.
• Create two equations: one for the sum and one for the product of the integers.
• Rearrange the equations if necessary to express one variable in terms of the other.

These principles guide students in finding the integer pairs that conform to specific conditions, enhancing their problem-solving skills.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, which means they have the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown. In the solved exercise, quadratic equations emerge when finding integers with specific sums and products. Here’s how it happens:

• Start with two equations based on given conditions: sum \(x + y = S\) and product \(xy = P\).
• Substitute one expression into the other to form a quadratic equation \(x^2 + bx + c = 0\) where \( b = -S\) and \( c = P\).

The solution to a quadratic equation can be found using various methods such as factoring, completing the square, or utilizing the quadratic formula. Understanding how to manipulate these equations is crucial for solving integer pair problems.

Factorization

Factorization is the process of decomposing a polynomial into a product of simpler polynomials. Every quadratic equation can be expressed as the product of its linear factors. This process is instrumental in solving quadratic equations, because finding the factors allows us to determine the roots of the equation easily.

• Transform the quadratic into the standard form \(ax^2 + bx + c = 0\).
• Identify factors of \(c\) that add up to \(b\).
• Write these as \((x + m)(x + n) = 0\) where \(m\) and \(n\) are factors.

The roots of the equation are simply the values of \(x\) that make each factor zero. Factorization efficiently simplifies the solving process, providing a straightforward pathway to find integer solutions.

Algebraic Expressions

Algebraic expressions lie at the heart of integer pair problems, consisting of variables raised to powers combined by arithmetic operations. In the exercise, the algebraic expressions become the backbone of the equations that need to be solved for \(x\) and \(y\). Here's how they function:

• An expression like \(xy = P\) evolves into quadratic equations through substitution.
• Expressions are manipulated to derive meanings based on problem requirements.

Algebraic expression manipulation demands understanding how these expressions relate to equations, particularly working with factors, sums, and simplified forms. Mastery of these expressions is essential for solving more complex algebra problems, benefiting overall mathematical proficiency.