Question

A positive real number is 2 less than another. If the sum of the squares of the two numbers is 58, find the numbers.

Answer 1

Answer:

x2 + (x+2)2 = 16

 

x2 + x2 + 4x + 4 = 16

 

2x2 + 4x - 12 = 0

 

x2 + 2x - 6 = 0

 

x = [-2 ± √28] / 2 = [-2 ± 2√7] / 2

 

                            = 2(-1 ± √7)/2 = -1 ± √7

 

-1 - √7 is negative, so x = -1 + √7 and x+2 = 1 + √7

Answer 2

Answer:

x = y - 2 ---(1)

x^2 + ( y- 2 )^2 = 58 ---(2)

substituting eq (1) in (2)

x^2 + x^2 = 58

2x^2 = 58

x^2 = 29

x = root 29