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The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?

Which of the following equations is the result of using the factoring method to solve the problem?
(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0

User Luissquall
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2 Answers

5 votes
Call n smallest number. Then n+1 is its consecutive.

n^2 + (n+1)^2 = 41
n^2 + n^2 + 2n + 1 = 41
2n^2 + 2n -40 = 0
n^2 + n -20 = 0
(n + 5)(n- 4) = 0

Then the answer is the third option.
User LavaScornedOven
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3 votes

Answer:

Factors are (n + 5)(n - 4) = 0 .

Explanation:

Given : The sum of the squares of 2 consecutive negative integers is 41. What are the numbers.

To find : Which of the following equations is the result of using the factoring method to solve the problem.

Solution : We have given statement

Let two consecutive number are : n and n +1 .

Square of two consecutive number are : n² and (n+1)².

According to question : sum of the squares of 2 consecutive negative integers is 41.

n² + (n+1)² = 41.

n² + n² + 1 +2n =41

2n² + 2n +1 =41

On subtracting 41 from both sides

2n² + 2n +1- 41 = 0

2n² + 2n - 40 = 0

On dividing by 2 to whole equation

n² + n - 20 = 0

On factoring

n² + 5n -4n - 20 = 0

Taking common n from first two terms and -4 from first two last terms

n (n +5) -4 (n +5) = 0

Grouping

(n -4) (n +5) = 0.

Therefore, Factors are (n + 5)(n - 4) = 0 .

User Thomasena
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