Question

The sum of the squares of three consecutive integer numbers is 1454. Find the numbers. I have already found 21 22 23 but the question says that there are 2 possibilities. please help!!!!!!

Answer 1

Answer:

three consecutive integers

Let x,(x+1),(x+2) represent the three consecutive integers

Question states

x^2 +(x+1)^2+(x+2)^2 = 110

Solving for x

3x^2 + 6x  + 5 = 110

3x^2 + 6x  -105 = 0

3(x^2 + 2x - 35)  = 0

 (x^2 + 2x - 35)  = 0  

factoring

 (x+7)(x-5) = 0 Note: SUM of the inner product(7x) and the outer product(-5x) = 2x

 (x+7)=0  |x = -7  The three consecutive integers are -5,-6,-7

 (x-5)=0  |x = 5   The three consecutive integers are 5,6,7

25 + 36 + 49 = 110  

Answer 2

The three consecutive integer numbers are 21, 22 and 23

Let the 3consecutive integers be x-1, x and x+1

Taking the sum of the squares

(x-1)²+x² + (x+1)² = 1454

x²-2x+1+x²+x²+2x+1 = 1454

3x²+2 = 1454

3x²= 1452

x² = 484

x = 22

First number is x - 1 = 22 - 1 = 21

second number is x = 22

Third number is x+ 1= 22+ 1 = 23

Hence the three consecutive integer numbers are 21, 22 and 23

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