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

Answer:

$a = -4 + 2\sqrt{6}, b = 4 + 2\sqrt{6}$

Step-by-step explanation:

Let a,b - the 2 numbers.

$a+8 = b\\a^2 + b^2 = 80\\a^2 + (a+8)^2 = 80\\2a^2 + 16a + 64 = 80\\2a^2 + 16a - 16 = 0\\a^2 + 8a - 8 = 0\\a_{12} = \frac{-8 \pm \sqrt{64 - 4(-8)}}{2} = \frac{-8 \pm 4\sqrt{6}}{2} = -4 \pm 2\sqrt{6}\\b_{12} = 4 \pm 2\sqrt{6}\\ 4 < 2\sqrt{6} => a = -4 + 2\sqrt{6}, b = 4 + 2\sqrt{6}$