Solving the Product of Two Consecutive Numbers: A Step-by-Step Guide

In mathematics, finding two consecutive numbers whose product equals a specific value is a classic problem that involves quadratic equations. Let's explore a detailed solution to the problem of finding two consecutive numbers whose product is 3422. This guide will cover various methodologies, including the quadratic formula and brute force approaches.

Method 1: Using the Quadratic Formula

Let the two consecutive numbers be ( n ) and ( n-1 ). The equation for their product can be set up as follows:

( n(n-1) 3422 )

This simplifies to:

( n^2 - n - 3422 0 )

Now we can use the quadratic formula to solve for ( n ):

( n frac{-b pm sqrt{b^2 - 4ac}}{2a} )

Here, ( a 1 ), ( b -1 ), and ( c -3422 ). Plugging in these values:

( n frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot -3422}}{2 cdot 1} )

( n frac{1 pm sqrt{1 13688}}{2} )

( n frac{1 pm sqrt{13689}}{2} )

( n frac{1 pm 117}{2} )

Calculating the two possible values for ( n ):

( n frac{116}{2} 58 )

( n frac{-118}{2} -59 ) (not applicable since we are looking for positive consecutive numbers)

Thus the two consecutive numbers are:

58 and 59

Verification

To verify:

( 58 times 59 3422 )

So the numbers are 58 and 59.

Method 2: Estimating Near Perfect Squares

We know that:

( nn - 1 n^2 - n ) and ( n^2 - n ) looks for the perfect squares near 4032.

We know that:

( 60^2 3600 ) and ( 21^2 4096 64^2 ) (which is a power of 2 and not very large).

The last digit of 4032 is 2, which is obtained when we multiply 2×1 or 4×3 or 7×6 or 9×8 as their last digit.

As 4032 is a bit smaller than 64×64, we try:

( 64 times 63 64^2 - 64 4096 - 64 4032 ).

Therefore, the answer is 63 and 64.

Method 3: Using J Programming Language (Brute Force Approach)

In J programming language:

( 0 1: sqrt{3422} )

The result is:

58 59

Factorization Method: Breaking Down the Factors

Factorizing 3422 2×2×855. We can make some trials and find the answer as 58×59 3422.

The numbers are:

( boxed{58, 59} )

Step-by-Step Analysis Using the Quadratic Formula

1. Let ( x ) be the first number.

Therefore, the second number is ( x - 1 ).

2. The product of the numbers is:

( x(x - 1) 3422 )

This simplifies to:

( x^2 - x - 3422 0 )

3. Rearrange the equation:

( x^2 - x - 3422 0 )

This quadratic expression is equivalent to:

( (x - 59)(x - 58) 0 )

The solutions to the quadratic equation are the two consecutive numbers being ( x 58 ) and ( x 59 ).

Thus, the numbers are:

( boxed{58 text{ and } 59} )

From this, we conclude that the numbers are 58 and 59, which fit the criteria of having a product of 3422.