# The average of 2 numbers is 12 and their product is 56. What is the average of their squares?

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Let the two numbers be A and B. The average of the two numbers is 12, therefore ( A + B) / 2 =12

=> A + B = 24

Their product is 56 or A*B = 56.

Now we know that (A+ B)^2 = A^2 + B^2 + 2AB.

The average of their squares is (A^2 + B^2) /2

= [( A + B)^2 - 2AB] / 2

= [ 24^2- 2* 56] / 2

= [ 576 - 112] /2

= 464/2

= 232

**The average of the squares of the two numbers is 232.**

Average of the 2 numbers is 12. Therefore their sum

= 2*12 = 24

Let the two numbers be represented by a and b.

Then:

a + b = 24 and

a*b = 56

We have to find the value of:

(a^2 + b^2)/2

We know:

(a + b)^2 = a^2 + b^2 + 2ab

Substituting in above equations known values of (a + b) and a*b:

24^2 = a^2 + b^2 + 2*56

==> 576 = a^2 + b^2 + 112

==> a^2 + b^2 = 576 - 112 = 464

Dividing both sides by 2:

(a^2 + b^2)2 = 232

Since the product of 2 numbers is 56, we pressume one number is x and the other number = 56/x.

So now the average of x and 56/x is (x+56/x)/2 is 12.

So (x+56/x)/2 = 12. Multiply both sides by 2x:

x^2 +56 = 24x

x^2 -24x+56= 0

Therefore sum of the roots: x1+x2 = -(-24) = 24 and their product x1x2 = 56

Therfore sum of their squares: x1^2+x2^2 = (x1+x2)^2-2(x1x2) = 24^2-2(56) = 464