SOLUTION: Find two real numbers that have a sum of 6 and a product of 4.

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Question 212506: Find two real numbers that have a sum of 6 and a product of 4.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find two real numbers that have a sum of 6 and a product of 4.

Step 1. Let x be one number

Step 2. Let y be the other number

Step 3. The problem statement gives us that x+y=6 or y=6-x

Step 4. We're also give that

xy=4

Substitute y=6-x

x%286-x%29=4

6x-x%5E2=4

Step 5. Now use above equation to obtain a quadratic one.

Add x%5E2-6x to both sides of the equation so left side will = 0

6x-x%5E2%2Bx%5E2-6x=4%2Bx%5E2-6x

0=x%5E2-6x%2B4


Step 6. So now we have a quadratic equation, so we can use the quadratic formula given as

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

where a=1, b=-6 and c=4 for our example. Follow steps below:


Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B4+=+0) has the following solutons:

x%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=%28-6%29%5E2-4%2A1%2A4=20.

Discriminant d=20 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+20+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+20+%29%29%2F2%5C1+=+5.23606797749979
x%5B2%5D+=+%28-%28-6%29-sqrt%28+20+%29%29%2F2%5C1+=+0.76393202250021

Quadratic expression 1x%5E2%2B-6x%2B4 can be factored:
1x%5E2%2B-6x%2B4+=+1%28x-5.23606797749979%29%2A%28x-0.76393202250021%29
Again, the answer is: 5.23606797749979, 0.76393202250021. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B4+%29




Step 7. ANSWER: The above steps give us two numbers : 5.2361 and 0.7639.


You can check to see if these numbers add up to 6 and their products equal 4 as given by the problem.

I hope the above steps were helpful. Good luck in your studies!

Respectfully,
Dr J

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