# Given the sum of two numbers = 3 and the product = 2 find the sum of the squares of the numbers.

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Let the two numbers we have to find the squares of be A and B.

Now their sum is A + B = 3

Their product is A*B = 2

We need A^2 + B^2.

This can be written as (A + B)^2 - 2A*B

=> 3^2 - 2*2

=> 9 - 4

=> 5

**Therefore the required sum of the squares is 5.**

The sum of the two numbers is 3. So we assume the numbers to be x and 3-x.. The product of the numbers is x(3-x) which is equal to 2. So x(3-x) = 2.

=> x*3-x^2= 2.

=> x^2-3x+2 = 0

=> (x-1)(x-2) = 0.

So x= 1 , or x = 2.

Therefore the two numbers are x = 1 and 3-x = 3-1 = 2. Or x = 2 and 3-x = 3-2 = 1.

So the assumed numbers are 1 and 2 , or 2 and 1.

Therefore the sum of the squares of the numbers = 1^2+2^2 = 1+4 = 5.

We'll note the sum as S and the product by P.

To determine the numbers, we'll form the quadratic equation and we'll determine it's roots.

x^2 - Sx + P = 0

S = 3 and P = 2

We'll substitute them into equation:

x^2 - 3x + 2 = 0

We'll apply the quadratic formula:

x1 =[3 + sqrt(9 - 8)]/2

x1 = (3+1)/2

x1 = 2

x2 = (3-1)/2

x2 = 1

**The numbers whose sum is 3 and product is 2 are: 1 and 2.**

**S = 1 + 2 = 3**

**P = 1*2 = 2**