Respuesta :

adebayodeborah8

Answer:

n = 21, 22, 23.

Step-by-step explanation:

let the 3 consequitive  numbers be  n, n +1 , n+2

fro the question,

let squares of three consecutive integer numbers be

n², (n+1)² , (n+2)²

let the sum of the squares of the three consecutive integer numbers  be

 n² + (n+1)² +(n+2)² =1454

by opening the bracket

we have;

n² + (n² + 2n +1) + (n²+ 2n + 4) =1454

n² + n² + 2n + 1 + n² +4n +4 = 1454

therefore;

3n² + 6n + 5 = 1454

3n² + 6n +5 - 1454 =0

3n² + 6n - 1449 = 0

divide through by 3

we have;

n² + 2n - 483 = 0

n² + 2n = 483

add 1 to both sides to make an equation

 n² + 2n + 1 = 483 +1

n (n +1) + 1 ( n+1) = 484

since the equation in the bracket is thesame, therefore we pick 1

n +1 (n +1) = 484

(n+1)² = 484

Square on both sides

n + 1 = √484

n +1 = 22

n =22 - 1

n = 21

since its three consecutive integers

and n = 21

therefore n + 1 = 21 + 1 = 22

N +2 = 21 + 2 = 23

Therefore n = 21, 22, 23.

adioabiola

The three consecutive numbers which have its sum of the square as 1454 is 21, 22 and 23

Let

the three consecutive numbers be

x

x x + 1

x x + 1x + 2

sum of their squares:

(x²) + (x + 1)² + (x + 2)² = 1454

x² + x² + 2x + 1 + x² + 4x + 4 = 1454

3x² + 6x + 5 = 1454

3x² + 6x + 5 - 1454 = 0

3x² + 6x - 1449 = 0

divide through by 3

x² + 2x - 483 = 0

x² + 2x = 483

add 1 to both sides

x² + 2x + 1 = 483 + 1

x² + 2x + 1 = 484

factorize

x(x + 1) + 1 (x + 1) = 484

(x + 1) (x + 1) = 484

(x + 1)² = 484

x + 1 = √484

x + 1 = 22

x = 22 - 1

x = 21

Therefore, the three consecutive numbers are

x = 21

x + 1

= 21 + 1

= 22

x + 2

= 21 + 2

= 23

Read more:

https://brainly.com/question/18867131

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