1. A positive real number is 6 less than another. If the sum of the squares of the two numbers is 38, then find the numbers.

Answer 1

x = y -6

x^2 + (x+6)^2 = 38

x^2 + x^2 + 12x + 36 = 38

2x^2 + 12x - 2 = 0

x^2 + 6x - 1 = 0

x = [-6 +or- sqrt( 36 - 4*1*-1)]/2

[-6 +or- sqrt(40)/ /2 =

(-6 + sqrt(40))/2 <--- only positive allowed, as given

(-6 + 2 * sqrt(10))/2

= -3 + sqrt(10)

which is APPROXIMATELY 0.16227766...

so the first number is sqrt(10 ) - 3 and the second number is sqrt(10) + 3

squaring them: 19 - 6*sqrt(10) and 19 + 6*Sqrt(10)

their total is 38