Find the whole number m for which (3 + √m)(3 - √m) = 4.

Step 1. Part (a). $(3 + \sqrt{m})(3 - \sqrt{m}) = 9 - m$. So $9 - m = 4$, and $m = 5$.

Step 2. Part (b). $\sqrt{20k} = 10 = \sqrt{100}$, so $20k = 100$, and $k = 5$.

Step 3. Check. $3 + \sqrt{5}$, $3 - \sqrt{5}$, $\sqrt{5}$ and $\sqrt{20} = 2\sqrt{5}$ are all irrational. Their products, $4$ and $10$, are rational.