Answer:
21
Step-by-step explanation:
Let the three consecutive integers be;
x , x + 1 and x + 2:
The sum of squares of the two smaller numbers:
The two smaller numbers are x and x+ 1
Sum of their squares:
x² + (x+1)²;
x² + x² + 2x + 1 = 2x² + 2x + 1
It is 45 more than the square of the larger number:
Larger number = x + 2
(x+2)² = x² + 4x + 4
45 more;
x² + 4x + 4 + 45 = x² + 4x + 49
Both are equal:
2x² + 2x + 1 = x² + 4x + 49
2x² - x² + 2x - 4x + 1 - 49 = 0
x² - 2x - 48 = 0
x² - 6x + 8x - 48 = 0
x(x - 6) + 8(x - 6) = 0
(x + 8)(x-6) = 0
x + 8 = 0 or x - 6 = 0
x = -8 or x = 6
We choose x = 6 because it makes the solution true as a positive integer
Sum of the three numbers:
x + x + 1 + x + 2 = 6 + 6 + 1 + 6 + 2 = 21
