## Introduction

Diamond math problems make it easy for us to understand the relation between sum and product of two numbers. In this article, you will learn about **diamond problems math** and how to do diamond problems manually and by using diamond math calculator. Let's start learning from here.

## How to do a Diamond Problem in Maths?

Before starting the process of doing diamond problems math, let's understand the formula first.

### Diamond Problem Formula

The diamond problem is a simple technique that helps the beginners to develop interest in maths. It is more helpful than the other mathematical techniques because its formula contains very simple rules.

In diamond math problem formula, there are two following rules;

SUM = A + B & Product = A × B

**Sum and Product rule**. The sum and product of two numbers is written in the top and bottom side of the diamond shape.**Unknown Numbers**. The unknown numbers A and B are written in the left and right side and they can be found by guessing technique.

## Steps of solving Diamond Problem

The diamond problem can be used by following some simple steps, these are

- Draw a diamond shape or a polygon or just a cross.
- Write the sum and product of two numbers in the top and bottom cells.
- Leave the left and write cells empty.
- Find the two numbers that result in the same given sum and product.

By using the above steps, we can easily do a diamond problem. See the following examples.

### Diamond math problems example no. 1

In this example we will learn how to find factors of a quadratic equation by doing diamond maths problems.

Consider the equation **15x ^{2} + 7x -2 = 0**

To find the factors of the above equation, we have to find **2** numbers that add or subtract up to **7** and multiplied to **30**.

So,

Sum = 7 and product = 30

Now, assuming the numbers by doing diamond problems.

sum | product |

7 | 30 |

10 | 3 |

Hence the numbers are **10** and **3**. They result in **7** when subtracted. So,

15x^{2} + 10x - 3x - 2 = 0

5x (3x + 2) -1 (3x + 2) = 0

(3x + 2) (5x - 1) = 0

**So the solution of given equation is**

x = { (-2) / 3 , 1 / 5}

### Diamond problems math example no. 2

In this example, we will learn how to find an unknown number by using the sum and product of two numbers.

Consider the sum and product of two numbers **(5 , x)** are **14** and **45**, where the number **x** is unknown.

Using diamond problem,

sum = 14 | product = 45 |

Number_{1} = 5 |
Number_{2} = x |

Since,

5 + x = 14 ⟹ x = 14 - 5

x = 9

Hence the unknown number is **9**. We can cross verify the result by using the product of these two numbers therefore,

5 × 9 = 45

### Diamond math problem example no. 3

In this example we will learn how to do diamond math problems with unknown numbers and unknown products.

Suppose the sum of two numbers **(3 , x)** is **27** and the product is known. So,

By using the sum relation,

3 + x = 27 ⟹ x = 27 -3 = 24

So,

sum = 27 | product = 72 |

3 | 24 |

## How to do a Diamond Problem Using a Calculator?

It is a more easy and smart way to find factors of numbers using the Diamond Problem Calculator. Follow the given steps to solve a problem with this tool.

- Enter the numbers in factor A and B boxes.
- Or use the load example button.
- Click on the calculate button.
- This tool will give you results within a few seconds.

## FAQ’s

### What is a Diamond Problem?

It is a skill developing technique that helps the beginners to learn maths by understanding the relation between sum and product of two numbers.

### How does Diamond Maths Work?

The diamond maths contains a rhombus shape divided in four sessions. The sum and product is written in the top and bottom cell of the diamond shape and left and write cells left empty to be filled by students.

### How do you solve diamond fraction problems?

Diamond problem with fraction is little bit different from the other numbers. For example if you have two numbers 2/-3 and 1/5, you can easily find the product by multiplying and sum by adding these fractions. Such as,

$$\sum i \frac{2}{3} + \frac{1}{5} \;=\; \frac{13}{15} $$

$$ Product \;=\; \frac{2}{3} \times \frac{1}{5} \;=\; \frac{2}{15} $$

### How do you do the box and diamond method?

It is easy to do box and diamond method for different problems like factoring the quadratic equations. For example, in the equation 2 x^{2}−5x + 3, you can use diamond problem.

sum = -5 | product = 6 |

N1 -3 | N2 -2 |

So the factors of the equations are,

2x^{2}−5x+3=2x^{2}−2x−3x+3

2x^{2}−5x+3=2x(x−1)−3(x−1)=(x−1)(2x−3)