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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Let's call the smaller number $x$.
According to the problem, the larger number is $x+4$.
The sum of the squares of the two numbers is $x^2+(x+4)^2$.
We are given that this sum is 72. So, we have the equation $x^2+(x+4)^2=72$.
Expanding the expression $(x+4)^2$ gives us $x^2+8x+16$.
So, our equation becomes $x^2+x^2+8x+16=72$.
Combining like terms, we have $2x^2+8x+16=72$.
Simplifying, we get $2x^2+8x-56=0$.
Dividing both sides by 2 gives us $x^2+4x-28=0$.
We can solve this quadratic equation by factoring. We look for two numbers whose product is -28 and whose sum is 4. The numbers are 7 and -4. Thus, we can factor the quadratic equation as $(x-4)(x+7)=0$.
Setting each factor equal to zero gives us $x-4=0$ or $x+7=0$. Solving each equation gives us $x=4$ or $x=-7$.
Since a positive real number was defined as $x$, $x=4$.
Therefore, the two numbers are $4$ and $4+4=\boxed{8}$.
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