Problem 92 - Algebra A Combined Function

Question

Fill in the chart by finding two numbers that have the given product and sum. The first column is filled in for you. $$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline & & \text { 85. } & \text { 86. } & \text { 87. } & \text { 88. } & \text { 89. } & \text { 90. } & \text { 91. } & \text { 92. } \\ \hline \text { Two Numbers } & 4,7 & & & & & & & & \\ \hline \text { Their Product } & 28 & 12 & 20 & 8 & 16 & -10 & -9 & -24 & -36 \\\ \hline \text { Their Sum } & 11 & 8 & 9 & -9 & -10 & 3 & 0 & -5 & -5 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Case 86: 2,6; Case 87: 4,5; Case 88: -1,-8; Case 89: -2,-8; Case 90: -2,5; Case 91: 3,-3; Case 92: 3,-8; Case 93: 4,-9.

Step 1: Understand the Problem

You need to fill in the chart by finding two numbers such that their product matches the given product in the chart and their sum matches the given sum in the chart for each case from 86 to 92.

Step 2: Equations Setup

For each case, solve the system of equations: if the two numbers are \( x \) and \( y \), then \( x \cdot y = \text{Product} \) and \( x + y = \text{Sum} \).

Step 3: Solve for Case 86

Given: Product = 12, Sum = 8. You need numbers \( x, y \) such that \( x \cdot y = 12 \) and \( x + y = 8 \). The numbers can be found by trial method: \( 2 \cdot 6 = 12 \) and \( 2 + 6 = 8 \). So, the numbers are 2 and 6.

Step 4: Solve for Case 87

Given: Product = 20, Sum = 9. Set up the equations \( x \cdot y = 20 \) and \( x + y = 9 \). For example, \( 4 \cdot 5 = 20 \) and \( 4 + 5 = 9 \). The numbers are 4 and 5.

Step 5: Solve for Case 88

Given: Product = 8, Sum = -9. Find two numbers such that \( x \cdot y = 8 \) and \( x + y = -9 \). The numbers \( -1 \) and \( -8 \) satisfy \( -1 \cdot (-8) = 8 \) and \( -1 + (-8) = -9 \). So, the numbers are -1 and -8.

Step 6: Solve for Case 89

Given: Product = 16, Sum = -10. Numbers must satisfy \( x \cdot y = 16 \) and \( x + y = -10 \). Two numbers that work are \( -2 \) and \( -8 \), since \( -2 \cdot (-8) = 16 \) and \( -2 + (-8) = -10 \). They are -2 and -8.

Step 7: Solve for Case 90

Given: Product = -10, Sum = 3. Numbers must satisfy \( x \cdot y = -10 \) and \( x + y = 3 \). Testing possibilities, the numbers \( -2 \) and \( 5 \) result in \( -2 \cdot 5 = -10 \) and \( -2 + 5 = 3 \). The numbers are -2 and 5.

Step 8: Solve for Case 91

Given: Product = -9, Sum = 0. The numbers must satisfy both \( x \cdot y = -9 \) and \( x + y = 0 \). The numbers 3 and -3 work because \( 3 \cdot (-3) = -9 \) and \( 3 + (-3) = 0 \). Numbers are 3 and -3.

Step 9: Solve for Case 92

Given: Product = -24, Sum = -5. Find numbers with \( x \cdot y = -24 \) and \( x + y = -5 \). The numbers \( 3 \) and \( -8 \) work since \( 3 \cdot (-8) = -24 \) and \( 3 + (-8) = -5 \). The numbers are 3 and -8.

Step 10: Solve for Last Case 93

Given: Product = -36, Sum = -5. Solve \( x \cdot y = -36 \) and \( x + y = -5 \). Numbers such as \( 4 \) and \( -9 \) give us \( 4 \cdot (-9) = -36 \) and \( 4 + (-9) = -5 \). The numbers are 4 and -9.

Key Concepts

system of equations

When solving problems like finding two numbers with a specific product and sum, we often use a system of equations. A system of equations involves solving two or more equations simultaneously. In this context, we are looking for two numbers, often denoted as \( x \) and \( y \). We have two key equations:

• \( x \cdot y = \text{Product} \)
• \( x + y = \text{Sum} \)

From these equations, we can deduce the values of \( x \) and \( y \) that satisfy both conditions. The system of equations offers a structured way to find the relationship between the product and sum of numbers. By solving the system, we can determine the exact numbers needed. This process highlights the interconnectedness of algebraic operations.

product and sum problems

Product and sum problems involve finding two numbers who multiply to a particular product and add up to a specific sum. These problems often require a strategic approach because they involve balancing two different operations. Initially, it might seem daunting to guess the right numbers, especially when negative values are part of the solution.

Here are a few strategies:

• List the factors of the product to see possible pairs.
• Check which pair adds up to the required sum.
• Be aware that sometimes negative numbers are involved, which might flip the expected sum or product sign.

These strategies highlight how interconnected multiplication (product) and addition (sum) are, which is a foundational element in algebra.

trial and error method

The trial and error method is an intuitive approach used to solve equations by testing various possible solutions. In essence, it involves systematically trying out different number pairs to see which meets both the given product and sum.

In our problem, this means:

• Select possible factors of the product.
• Test each factor pair to check if it also sums up to the required value.
• If a pair does not satisfy both conditions, move on to another pair.

This method can be time-consuming, but it develops problem-solving instincts and is often used when algebraic manipulation becomes complex or when the numbers involved are small.

solving quadratic equations

In algebra, sometimes solving a system of equations leads to a quadratic equation maneuver. For certain product and sum problems, especially when no clear solution appears through trial, recognizing a quadratic pattern can be useful. The general form of a quadratic equation is:

• \( ax^2 + bx + c = 0 \)

For our sum and product problem, we can represent it by setting \( x \) and \( y \) such that:

• \( x + y = b \, (\text{where } b \text{ is the sum}) \)
• \( xy = c \, (\text{where } c \text{ is the product}) \)

Reframing it in terms of a quadratic, we arrive at:

• \( x^2 - bx + c = 0 \)

Solving this quadratic gives precise values for \( x \) and \( y \), and knowing how to manipulate within this framework lends accuracy and depth, especially when conjecturing fails.