# 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.**

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.