The sum of the squares of two consecutive positive integers is 41. Find the integers

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