Question 27: Find a natural number x so that \( D = \frac{{\sqrt x – 3}}{{\sqrt x + 2}}\) has an integer value.

\( D = \frac{{\sqrt x – 3}}{{\sqrt x + 2}}\)

To D∈Z then \( \left( {\sqrt x + 2} \right)\) must be in Z and be a divisor of 5.

Because \( \left( {\sqrt x + 2} \right) > 0\) so there are only two cases:

Case 1:

\( \sqrt x + 2 = 1 \Leftrightarrow \sqrt x = – 1(vl)\)

Case 2:

\( \sqrt x + 2 = 5 \Leftrightarrow \sqrt x = 3 \Leftrightarrow x = 9\) (satisfied)

So let

D∈Z then x=9 (then D=0).

The answer to choose is: