Ans: B
Let $x$ and $y$ be the two numbers with $x>y$. Hence, we have
Let $x$ and $y$ be the two numbers with $x>y$. Hence, we have
$\left\{ \begin{array}{l}
x+y=34 \\
xy = 120
\end{array} \right.$
For $x>y$, we have
$\begin{array}{rcl}
(x-y)^2 & = & x^2 -2xy +y^2 \\
& = & x^2 +2xy + y^2 -4xy \\
& = & (x+y)^2 -4xy \\
& = & (34)^2 -4(120) \\
& = & 676 \\
x-y & = & \sqrt{676} \\
& = & 26
\end{array}$
Therefore, the difference between the two numbers is $26$.
