Answer: 16 and 25
Step-by-step explanation:
Let the consecutive terms = x² + ( x + 1 )²
Since their sum = 41
we now open the bracket and equate it to 41
x² + x² + 2x + 1 = 41
2x² + 2x + 1 = 41
Re arranged and solve quadractically
2x² + 2x + 1 - 41 = 0
2x² + 2x - 40 = 0
Reduce to lowest term to easy solving by dividing by 2
x² + x - 20 = 0
solve x² + 5x - 4x - 20 = 0
x( x + 5 ) - 4( x + 5 ) = 0
consider common factors
( x + 5 )( x - 4 ) = 0
when x + 5 = 0 , x = -5 and when x - 4 = 0 , x = 4
Therefore
x = -5 or 4
The integers x² and (x + 1)²
Now substitute for the numbers
When x = -5,
(x)² = (-5)²
= 25 and when
(x + 1)² = (-5 + 1)²
= (-4)²
= 16.
When x = 4
x² = (4)²
= 16
(x + 1)²= (4 + 1)²
= 5²
= 25
So the numbers are 16 and 25
