Question

a positive real number is 8 less than another. if the sum of the squares of the two numbers is 80 find the numbers

Answer 1

`a = -4 + 2√(6), b = 4 + 2√(6)`

Explanation:

Let a,b - the 2 numbers.

`a+8 = b\\a^2 + b^2 = 80\\a^2 + (a+8)^2 = 80\\2a^2 + 16a + 64 = 80\\2a^2 + 16a - 16 = 0\\a^2 + 8a - 8 = 0\\a_(12) = (-8 \pm √(64 - 4(-8)))/(2) = (-8 \pm 4√(6))/(2) = -4 \pm 2√(6)\\b_(12) = 4 \pm 2√(6)\\ 4 < 2√(6) => a = -4 + 2√(6), b = 4 + 2√(6)`