Let's assume the two numbers are x and y. We are given two pieces of information:
1) The square of the sum of the two numbers is 144: (x + y)^2 = 144
Expanding this equation, we get x^2 + 2xy + y^2 = 144.
2) The sum of their squares is 80: x^2 + y^2 = 80.
We can solve this system of equations by subtracting equation 2 from equation 1 to eliminate the y^2 term:
x^2 + 2xy + y^2 - (x^2 + y^2) = 144 - 80
2xy = 64
xy = 32
Now let's use this value of xy in either equation 1 or equation 2 to find the values of x and y.
Let's use equation 2:
x^2 + y^2 = 80
x^2 + (32/x)^2 = 80
Rearranging this equation, we get x^4 - 80x^2 + 1024 = 0.
We can factor this equation as (x^2 - 64)(x^2 - 16) = 0.
This gives us two equations:
1) x^2 - 64 = 0
x^2 = 64
x = ±8
2) x^2 - 16 = 0
x^2 = 16
x = ±4
So the numbers are x = 8, y = 4 or x = -8, y = -4.
The square of the sum of two numbers is 144, and the sum of their squares is 80. Find the number
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