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.
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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
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