A positive real number is 4 more than another. If the sum of the squares of the two numbers is 44, find the numbers.

Question

write your answers in the form p +/-q square root r

Answer 1

X=Y+4

x^2 + y^2 = 44

(y+4)^2 + y^2 = 44

y^2 + 8y + 16 + y^2 = 44

2y^2 + 8y - 28 = 0

y^2 + 4y - 14 = 0

y = [-4 +or- sqrt( 16 + 56)] / 2 = [-4 +or- sqrt(72)]/2 = [-4 +or- 6*sqrt(2)]/2 = -2 +or- 3*Sqrt(2)

for y = -2 + 3*sqrt(2), then x = y+4 = 2 + 3 * sqrt(2)

The sum of the squares is 4 - 12*Sqrt(2) + 18 + 4 + 12*Sqrt(2) + 18 = 44

for y = -2 - 3 * sqrt(2) , then x = y+4 = 2 + 3 * sqrt(2)

the sum of the square is 4 -12 * sqrt(2) + 18 + 4 + 12 * sqrt(2) + 18= 44