SOLUTION: The product of two positive consecutive numbers is 42. What is the larger number?

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Question 225878: The product of two positive consecutive numbers is 42. What is the larger number?
Found 2 solutions by Alan3354, drj:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by drj(1380) About Me  (Show Source):
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The product of two positive consecutive numbers is 42. What is the larger number?

Step 1. Let n be the smaller positive number.

Step 2. Let n+1 be the next consecutive and larger positive number.

Step 3. Then n(n+1)=42 since their product is 42.

Step 4. Subtract 42 from both sides of the equation to get a quadratic.

n%5E2%2Bn-42=0

Step 5. To solve, use the quadratic equation given below as

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=1, b=1, and c=-42

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



Selecting the positive solution n=6, then n+1=7. Also note their product is 42.

Step 6. ANSWER: The larger number is 7.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV