SOLUTION: The sum of two numbers is 12 ans the sum of their squares is 78. Find the numbers. I do not believe that either number is whole.

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Question 95825: The sum of two numbers is 12 ans the sum of their squares is 78. Find the numbers.
I do not believe that either number is whole.
Found 2 solutions by checkley71, Edwin McCravy:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
X+Y=12 OR X=12-Y
X^2+Y^2=78
(12-Y)^2+Y^2=78
144-24Y+Y^2+Y^2=78
2Y^2-24Y+144-78=0
2Y^2-24Y+66=0
2(Y^2-12Y+33)=0
USING THE QUADRATIC EQUATION YOU ARE CORRECT:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(12+-SQRT[-12^2-4*1*33])/2*1
X=(12+-SQRT[144-132])/2
X=(12+-SQRT[12])/2
X=(12+-3.464)/2
X=(12+3.464)/2
X=15.464/2
X=7.73 ANSWER.
X=(12-3.464)/2
X=8.536/2
X=4.27 ANSWER.
PROOF
7.73+4.27=12
12=12
7.73^2+4.27^2=78
59.7529+18.2329=78
78=78

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

The sum of two numbers is 12 and the sum of their 
squares is 78. Find the numbers. 

I do not believe that either number is whole.

--------------------------------------------
Let x = one number
Let y = other number

            x  + y  = 12
            x² + y² = 78

Solve the first equation for y

                  y = 12-x

Substitute (12-x) for y in the second equation:

            x² + y² = 78
       x² + (12-x)² = 78
x² + 144 - 24x + x² = 78

2x² - 24x + 66 = 0

x² - 12x + 33 = 0
         _
x = 6 ± Ö3
              _                    _       _
When x = 6 + Ö3, y = 12-x = 12-(6+Ö3) = 6-Ö3 

So one solution is 
             _     _
(x, y) = (6+Ö3, 6-Ö3)

              _                    _       _
When x = 6 - Ö3, y = 12-x = 12-(6-Ö3) = 6+Ö3 

So the other solution is 
             _     _
(x, y) = (6-Ö3, 6+Ö3)

Edwin