SOLUTION: A positive real number is 2 less than another. If the sum of the squares of the two numbers is 6, find the numbers.

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Question 1199231: A positive real number is 2 less than another. If the sum of the squares of the two numbers is 6, find the numbers.
Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x be the smaller positive real number.


Then the larger number is (x+2), and the problem leads to equation

    x^2 + (x+2)^2 = 6    (the sum of squares is 6).


Simplify and find x

    x^2 + (x^2 + 4x + 4) = 6

    2x^2 + 4x -2 = 0

     x^2 + 2x - 1 = 0


Use the quadratic formula

     x%5B1%2C2%5D = %28%28-2%29+%2B-+sqrt%282%5E2+-+4%2A1%2A%28-1%29%29%29%2F2 = %28-2+%2B-+sqrt%288%29%29%2F2 = -1+%2B-+sqrt%282%29.


ANSWER.  The numbers are  x = -1%2Bsqrt%282%29  (the positive root) and 1%2Bsqrt%282%29.

Solved.