What is the sum of the squares of two numbers, the sum of which is 14 and their product is 8?
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Let the numbers to be determined be A and B.
Now the sum of the numbers is 14
=> A + B = 14.
The product of the numbers is 8
=> AB = 8.
We need to find the sum of the product of these numbers, for this we use the formula
(a + b)^2 = a^2 + b^2 + 2ab
=> a^2 + b^2 = (a + b)^2 - 2ab
A^2 + B^2 = (A+B)^2 – 2AB
=> 14^2 – 2*8
=> 196 – 16
=> 180
Therefore the sum of the squares of the two numbers is 180.
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If we know the sum and the product, we can form the quadratic:
x^2 - 14x + 8 = 0
Now, we'll consider the numbers that verify the quadratic as x1 and x2:
x1^2 - 14x1 + 8 = 0 (1)
x2^2 - 14x2 + 8 = 0 (2)
We'll add (1) and (2):
x1^2 + x2^2 = 14(x1 + x2) - 16
But the sum x1 + x2 = 14
x1^2 + x2^2 = 14*14 - 16
x1^2 + x2^2 = 196 - 16
x1^2 + x2^2 = 180
Let two numbers be x1 and x2. Their sum = x1+x2 and their product = x1x2 = 8.
Therefore x1 and x2 are the roots of the equation x^2-14x+8 = 0.
x^2 = 14x-8.......(1).
Substitute the roots x1 and x2 in (1).
x1^2 = 14x1-8........(2)
x2^2 = 14x2^2-8....(3)
(2)+(3): x1^2+x2^2 = 14(x1+x2) -16.
=> x1^2+x2^2 = 14(14)-16 = 180.
Therefore the sum of the squares of the numbers is 180.
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