Question

The sum of the squares of two consecutive integers is 41. find the integers

Answer 1

≡ We know that:
⇔ Consider the first integer is
`a`
⇔ So, the next integer is
`(a+1)`

≡ Solution:
⇒
`(a)^(2)+(a+1)^(2)=41`
⇒
`a^(2)+(a^(2)+2a+1)=41`
⇒
`2a^(2)+2a-40=0`
⇒
`(2a-8)(a+5)`
⇒
`a_(1)=(8)/(2)=\boxed{4}`║
`a_(2)=-5`

∴ So, there are 2 options,
`\boxed{4}and\boxed{5}` or
`\boxed{-5}and\boxed{-4}`

Answer 2

Start by declaring variables.
Since we have two consecutive integers, let your variables be x and x + 1.

Thus, we can say that:
x² + (x + 1)² = 41
x² + x² + 2x + 1 = 41
2x² + 2x - 40 = 0
x² + x - 20 = 0
(x + 5)(x - 4) = 0

So, x = -5 or 4.

Thus, we can have -5 and -4,
or 4 and 5.