The sum of the squares of three consecutive integer numbers is 1454. Find the numbers.

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

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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.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation: