SOLUTION: a positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.

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Question 1163433: a positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
system%28x-4%2Cand%2Cx%5E2%2B%28x-4%29%5E2=72%29

x%5E2%2Bx%5E2-8x%2B16=72

2x%5E2-8x=56
x%5E2-4x=28
cross%28x%5E2-4x%2B4=56%2B4%29-----------------WRONG.
(the rest, removed)

Answer by ikleyn(52815) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by @josgarithmetic has an error.

            I came to bring a correct solution. 


Let x be the smaller number; then the greater number is (x+4).


You have this equation

    x^2 + (x+4)^2 = 72.


Simplify and solve using the quadratic formula

    x^2 + x^2 + 8x + 16 = 72

    2x^2 + 8x - 56 = 0

    x^2  + 4x - 28 = 0

    x%5B1%2C2%5D = %28-4+%2B-+sqrt%284%5E2+%2B4%2A28%29%29%2F2 = %28-4+%2B-+sqrt%28128%29%29%2F2 = %28-4+%2B-+8%2Asqrt%282%29%29%2F2 = -2+%2B-+4%2Asqrt%282%29%29.


So, the numbers are  x%5B1%5D = -2+%2B+4%2Asqrt%282%29  and  x%5B2%5D = -2+-+4%2Asqrt%282%29.      ANSWER

Solved and completed.