Do you include the "10 H2O" in the formula for borax?
You are talking about the waters of hydration in crystal structure $\ce{Na2B4O7 \cdot 10H2O}$, and the answer is invariably—yes.

These water molecules account for 47.24% of the crystals weight! If you wanted to make $1\: \mathrm{L}$ of $1\: \mathrm{M}$ of $\ce{Na2B4O7}$, and you did not account for these waters of hydration, you would end up making a $0.4818\: \mathrm{M}$ solution. Less than half the desired concentration! 

This is the result of a double effect. Not only are you not adding as much salt as you thought you were, but because the salt contains water it is diluting the solution even further. In the case above $95.05\: \mathrm{mL}$ of water are actually added to the solution just in the salt!