A very common exam phrasing goes like this: "Show that (3 + \sqrt{2})^2 + (3 - \sqrt{2})^2 is a rational number." Or this: "Express (1 + \sqrt{5})^3 in the form p + q\sqrt{5} and find p and q." Or even just: "Simplify (2 + \sqrt{3})(2 - \sqrt{3})."
All three problems look different, but the habit that solves them is the same. You separate the expression into two groups — the rational (non-surd) terms and the surd terms — and then you handle each group on its own. If the problem promised rational, the surds should cancel cleanly. If the problem asked for p + q\sqrt{r}, the grouping hands you p and q directly.
This article is about training that grouping reflex.
The habit
Whenever you see an expression that mixes integers and surds, drill this four-step routine:
- Expand everything fully. Open every bracket, distribute every product.
- Collect rational terms — every piece that has no surd in it.
- Collect surd terms — every piece that has a \sqrt{\,\,} in it, keeping its coefficient.
- Simplify each group independently.
The final expression is just the sum of the two simplified groups. That is it. Once you drill this, every "show this is rational" problem becomes mechanical.
Worked example 1 — the form p + q\sqrt{r}
Simplify (2 + \sqrt{3})^2 - (2 - \sqrt{3})^2.
Expand each square using (a \pm b)^2 = a^2 \pm 2ab + b^2:
Subtract:
Rational group: 7 - 7 = 0. Surd group: 4\sqrt{3} + 4\sqrt{3} = 8\sqrt{3}.
Answer: 8\sqrt{3}. The rational part is zero — the whole thing is a pure surd.
Worked example 2 — show that X is rational
Show that (3 + \sqrt{2})^2 + (3 - \sqrt{2})^2 = 22.
Expand:
Add:
Surd group: +6\sqrt{2} - 6\sqrt{2} = 0. Good — the surds cancel, exactly as a "rational answer" problem promises. Rational group: 11 + 11 = 22.
Answer: 22, a rational number. The cancellation of the surd group is the proof.
Worked example 3 — a product that is rational
Show that (2 + \sqrt{3})(2 - \sqrt{3}) = 1.
The fast way is the conjugate identity (a + b)(a - b) = a^2 - b^2:
But it is worth doing it the long way once, because it shows the grouping in action. Distribute fully:
Rational group: 4 - 3 = 1. Surd group: -2\sqrt{3} + 2\sqrt{3} = 0.
Answer: 1. The cross terms — the ones carrying the surd — always cancel when you multiply conjugates. This is why rationalising denominators works.
Worked example 4 — solving for p and q
Given (1 + \sqrt{5})^3 = p + q\sqrt{5}, find p and q.
Use the binomial expansion (a+b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3 with a = 1 and b = \sqrt{5}:
Rational group: 1 + 15 = 16. Surd group: 3\sqrt{5} + 5\sqrt{5} = 8\sqrt{5}.
So (1 + \sqrt{5})^3 = 16 + 8\sqrt{5}, which means p = 16 and q = 8.
The grouping literally hands you the answer. No guessing, no matching coefficients by eye.
Why grouping works
Here is the conceptual reason the habit is bulletproof. The rational numbers and the surd multiples (like \sqrt{3}, \sqrt{5}, \sqrt{6}) live in different algebraic "slots." Under addition, they do not mix — you cannot add 3 and 2\sqrt{5} into a single simpler number, any more than you can add apples and kilometres.
The only way they interact is through multiplication. When you multiply (2 + \sqrt{3}) by (\sqrt{3} + 1), the \sqrt{3} \cdot \sqrt{3} becomes 3 (rational), and everything else stays where it was. But once you have finished expanding, each term is locked into one slot forever — rational, or some specific surd. Grouping is just the act of looking at each term and asking, "Which slot are you in?"
Multiple surd types — group each separately
What if you have more than one kind of surd in the expression? Simplify
Rational group: 5 - 7 = -2. \sqrt{3}-group: 2\sqrt{3} + 4\sqrt{3} = 6\sqrt{3}. \sqrt{2}-group: 3\sqrt{2}.
Answer: -2 + 6\sqrt{3} + 3\sqrt{2}.
You cannot combine 6\sqrt{3} and 3\sqrt{2} into anything simpler. They are different surds — different slots. Treat each surd type as its own independent group, just as you treat x-terms and y-terms separately in algebra.
When the surds do not cancel and they should
If a problem promised a rational answer and your final expression still carries a surd, stop. Do not force an answer. Retrace. The surd that refuses to vanish is the problem telling you a mistake happened upstream. Common causes:
- Sign error on the -b side of a conjugate expansion.
- Dropped coefficient — you wrote 2\sqrt{3} where you meant 4\sqrt{3}.
- Mis-expanded square — forgot the middle 2ab term of (a+b)^2.
- Forgot that (\sqrt{5})^2 = 5, not \sqrt{25} left as a surd.
The "surds should cancel" check is a free error detector. Use it.
Using groupings to prove rationality
Three patterns come up over and over in proof questions. Learn to spot them:
- Sum of conjugates: (a + b\sqrt{r}) + (a - b\sqrt{r}) = 2a. The surd group is +b\sqrt{r} - b\sqrt{r} = 0. The sum is rational.
- Product of conjugates: (a + b\sqrt{r})(a - b\sqrt{r}) = a^2 - b^2 r. Difference of squares — rational.
- Difference of conjugates: (a + b\sqrt{r}) - (a - b\sqrt{r}) = 2b\sqrt{r}. The rational group cancels; what is left is a pure surd.
If a problem hands you an expression built from a conjugate pair, you can usually predict the answer's shape before expanding.
Worked example 5 — olympiad-ish
Show that (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 1.
Difference of squares: 7^2 - (4\sqrt{3})^2 = 49 - 16 \cdot 3 = 49 - 48 = 1. Rational, as promised.
Or the long way: distribute fully, get 49 - 28\sqrt{3} + 28\sqrt{3} - 48, and watch the \sqrt{3}-group -28\sqrt{3} + 28\sqrt{3} evaporate. Rational group: 49 - 48 = 1. Same answer.
Recognition drill
For each expression, separate into rational and surd groups before writing the final form.
- (\sqrt{5} + 3)^2 = 5 + 6\sqrt{5} + 9. Rational: 14. Surd: 6\sqrt{5}. Answer: 14 + 6\sqrt{5}.
- (\sqrt{2} - 1)^2 + (\sqrt{2} + 1)^2 = (2 - 2\sqrt{2} + 1) + (2 + 2\sqrt{2} + 1). Rational: 6. Surd: 0. Answer: 6 (rational).
- (2\sqrt{3} + 1)(\sqrt{3} - 1) = 2 \cdot 3 - 2\sqrt{3} + \sqrt{3} - 1 = 6 - \sqrt{3} - 1. Rational: 5. Surd: -\sqrt{3}. Answer: 5 - \sqrt{3}.
Do five of these a day for a week and the habit burns in.
Closing
Surd terms and rational terms live in different algebraic slots. They add within their own slot; they only cross slots through multiplication. Once expansion is done, every term is locked into one group — and grouping them explicitly turns every "mixed surd + integer" problem into a mechanical two-column sort.
Two bonuses come free with the habit. If the problem promised a rational answer, a leftover surd is an instant flag that you made a computation error. And if the problem asked for p + q\sqrt{r}, the two groups are literally p and q\sqrt{r} — no matching, no guessing. The grouping is the answer.