If a > b are two real numbers such that a + b = 10 and a² + b² = 52, t

Now we need to find the value of $a-b$. We know the identity $(a-b)^2 = a^2 – 2ab + b^2$.
We can rewrite this as $(a-b)^2 = (a^2 + b^2) – 2ab$.
Substitute the values of $a^2+b^2$ and $ab$ that we found:
$(a-b)^2 = 52 – 2(24)$
$(a-b)^2 = 52 – 48$
$(a-b)^2 = 4$.
Taking the square root of both sides:
$a-b = \pm \sqrt{4} = \pm 2$.
The problem states that $a > b$, which means $a-b$ must be a positive value.
Therefore, $a-b = 2$.