SOLUTION: Find 2 numbers m and n whose sum is 10 and whose product is 18.

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Question 227053: Find 2 numbers m and n whose sum is 10 and whose product is 18.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find 2 numbers m and n whose sum is 10 and whose product is 18.

Step 1. Let m+n=10 since the sum is 10.

Step 2. Let mn=18 since the product is 18.

Step 3. Then m=10-n and n(10-n)=18. Subtract 18 from both sides of the equation leads to

10n-n%5E2-18=0

or

n%5E2-10n%2B18=0

Step 4. To solve use the quadratic equation given as

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

where a=1, b=-10, and c=18.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 1n%5E2%2B-10n%2B18+=+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=%28-10%29%5E2-4%2A1%2A18=28.

Discriminant d=28 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+28+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+28+%29%29%2F2%5C1+=+7.64575131106459
n%5B2%5D+=+%28-%28-10%29-sqrt%28+28+%29%29%2F2%5C1+=+2.35424868893541

Quadratic expression 1n%5E2%2B-10n%2B18 can be factored:
1n%5E2%2B-10n%2B18+=+1%28n-7.64575131106459%29%2A%28n-2.35424868893541%29
Again, the answer is: 7.64575131106459, 2.35424868893541. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-10%2Ax%2B18+%29



Note the above solutions in n are the numbers whose sum is 10 and product is 18.

Step 5. ANSWER: Then numbers are approximately 2.35 and 7.65.

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