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

User Birderic
by
7.6k points

2 Answers

1 vote
≡ 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}
User Alexey Soshin
by
7.4k points
6 votes
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.
User Xirururu
by
7.9k points

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