SOLUTION: (pretend exponent ^4) 3. The product of two consecutive integers, n and n + 1, is 42. What is the positive integer that satisfies the situation?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: (pretend exponent ^4) 3. The product of two consecutive integers, n and n + 1, is 42. What is the positive integer that satisfies the situation?       Log On


   

Question 950439: (pretend exponent ^4)
3. The product of two consecutive integers, n and n + 1, is 42. What is the
positive integer that satisfies the situation?

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
n(n+1)=42
n^2+n=42
n^2+n-42=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 1n%5E2%2B1n%2B-42+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-42=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+169+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%281%29%2Bsqrt%28+169+%29%29%2F2%5C1+=+6
n%5B2%5D+=+%28-%281%29-sqrt%28+169+%29%29%2F2%5C1+=+-7

Quadratic expression 1n%5E2%2B1n%2B-42 can be factored:
1n%5E2%2B1n%2B-42+=+1%28n-6%29%2A%28n--7%29
Again, the answer is: 6, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-42+%29

The positive answer is 6.
ANSWER: The positive integer that satisfies the situation is 6.