You can
put this solution on YOUR website!The product of two consecutive odd integers is 1 less than four times their sum.
Find the two integers.
Step 1. Let n be one odd integer.
Step 2. Let n+2 be the next consecutive odd integer.
Step 3. Translate problem statement into an equation such than n(n+2)=4(n+n+2)-1 since product is 1 less than four times their sum
where n(n+2) is the product of the two consecutive integers and (n+n+2) is the sum of the two integers
Step 4. Simplify to obtain a quadratic equation. Subtract 8n+7 from both sides of equation to get the right side of equation to be zero.
Step 5. Now we have a quadratic equation. So use quadratic formula given as
where
a=1, b=-6 and c=-7
The following steps solves the quadratic equation
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=64 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 7, -1.
Here's your graph:
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Step 6. ANSWER: Based on the above steps, choose 7 since it is positive. The integers are then 7 and 9.
Step 7. Verify the answer using n(n+2)=4(n+n+2)+1 in Step 3.
So it checks out.
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra.
