consecutive numbers

Question

The square of the sum of two consecutive natural numbers is greater than the sum of the squares of these two numbers by 112. Find the two numbers.

The square of the sum of two consecutive natural numbers is greater than the sum of the squares of these two numbers by 112. Find the two numbers.

Answer 1

 

I did this problem for you before.

Why do you want to hear it again?

 

First natural number is X where X can be 0,1,2,....

The next consecutive one is X+1

 

The square of the sum of two consecutive natural numbers:

 

 [ X + X+1]^2 = (2x+1)^2 = 4x^2 + 4x +1

 

The sum of the squares of these two numbers:

   X^2 + (x+1)^2 = x^2 + x^2 + 2x + 1 = 2x^2 + 2x + 1

 

They differ by 112. The former expression is 112 greater than the latter expression

 

4x^2 + 4x +1 - (2x^2 + 2x + 1) = 112

 

4x^2 + 4x +1 - 2x^2 - 2x - 1 = 112

 

2x^2 + 2x = 112

 

x^2 + x = 56 <--- divides everything by 2

 

x^2 + x - 56 = 0  <-- writes the standard quadratic

 

(x   +   8)(x   -     7 ) = 0

 

By zero product property, X = -8 or X = 7

 

Since -8 is not a natural number it is disqualified.

 

So the square of the sum of 7 and 8 is (7+8)^2 = 15^2 = 225

The sum of the squares of 7 and 8 is 7^2 + 8^2 = 49 + 64 = 113

They differ by 112.

 

Answer 2

Let the first natural number be N. Then the consecutive natural number is N+1

So the natural numbers are n and n+1, where n is an integer greater than -1.

 

 

(n + n + 1)^2 = n^2 + (n+1)^2 + 112

(2n + 1)^2 = n^2 + (n+1)^2 + 112

4n^2 + 4n + 1 = n^2 + n^2 + 2n+ 1 + 112

4n^2 + 4n + 1 = 2n^2 + 2n + 113

2n^2 + 2n - 112 = 0

n^2 + n - 56 = 0

(n + 8) ( n - 7) = 0

n = 7 or n = -8

n=7 --> (7 + 8)^2 = 15^2 = 225

7^2 + 8^2 = 49 + 64 = 113

They differ by 112

Answer 3

Hi Kiranpreet!

 

Two consecutive numbers can be represented by n and n+1.

 

The "square of the sum of two consecutive numbers" is represented by

 

(n + n+1)^2 = (2n+1)^2 = 4n^2 + 4n + 1

 

The "sum of the squares of these two numbers" is represented by

 

n^2 + (n+1)^2 = n^2 + (n^2 + 2n + 1) = 2n^2 + 2n + 1

 

The difference of these two expressions is 112.  So,

 

(4n^2 + 4n + 1) - (2n^2 + 2n + 1) = 112

4n^2 + 4n + 1 -2n^2 - 2n - 1 = 112

2n^2 + 2n = 112

2n^2 + 2n - 112 = 0

 

Divide both sides by 2:

 

n^2 + n - 56 = 0

(n + 8)(n - 7) = 0

 

So either n=-8  or  n=7

 

-8 is not a natural number, so the first number is 7.  The second number is then 8.

 

Hope this helps.