Conjugate Product: (√a + √b)(√a − √b) Jumps to a − b

Here is a specific recognition skill that saves about thirty seconds per problem in a JEE paper, and cumulatively several minutes across a section on surds. Whenever you see the product

(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})

do not expand it term by term. Do not write out four products. Do not distribute. Just jump straight to the answer:

(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b.

That is the whole move. Spot the shape, write the answer, move on.

The trigger

The shape is two factors of the form

(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})

or any permutation — (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) is the same thing, and so is (\sqrt{b} + \sqrt{a})(\sqrt{a} - \sqrt{b}) after reordering. The key signature is same two radicals, opposite signs between them. The moment you see that, the answer is a - b — the radicand of the first minus the radicand of the second (or whichever way the signs specify).

Examples the trigger fires on:

(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) jumps to 7 - 3 = 4.
(\sqrt{11} - \sqrt{5})(\sqrt{11} + \sqrt{5}) jumps to 11 - 5 = 6.
(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) jumps to x - y.

Examples where the trigger does not apply:

(\sqrt{5} + \sqrt{3})^2 — same sign, not opposite. This is a perfect square, and expanding gives 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}. Different pattern.
(\sqrt{5} + \sqrt{3})(\sqrt{7} - \sqrt{3}) — two different radicals on the first factor vs the second. Expand normally.
(\sqrt{5} + 2)(\sqrt{5} - 2) — but wait, this does fire, with a = 5 and b = 4 (since 2 = \sqrt{4}). The answer is 5 - 4 = 1. Worth recognising the "one radical, one integer" variant.

Why the pattern collapses

The identity is just the difference of squares (x + y)(x - y) = x^2 - y^2 applied with x = \sqrt{a} and y = \sqrt{b}:

(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b.

Why the cross terms vanish: a term-by-term expansion would give \sqrt{a}\cdot\sqrt{a} - \sqrt{a}\cdot\sqrt{b} + \sqrt{b}\cdot\sqrt{a} - \sqrt{b}\cdot\sqrt{b} = a - \sqrt{ab} + \sqrt{ab} - b = a - b. The two \sqrt{ab} cross terms cancel exactly because one has a plus and the other a minus — that is what the difference-of-squares pattern guarantees.

So the shortcut is not a magic rule — it is just the difference-of-squares identity spotted in its disguised form. But the disguise is so common that learning to see through it is genuinely useful.

The general form with coefficients

Slightly more general pattern: coefficients on the radicals.

(p + q\sqrt{r})(p - q\sqrt{r}) = p^2 - q^2 r.

Here p could be a rational number and q\sqrt{r} is a surd. The same difference-of-squares collapse applies. Example: (2 + 3\sqrt{5})(2 - 3\sqrt{5}) = 4 - 45 = -41.

Another variant: both sides have coefficients.

(p\sqrt{a} + q\sqrt{b})(p\sqrt{a} - q\sqrt{b}) = p^2 a - q^2 b.

Example: (3\sqrt{2} + 5\sqrt{3})(3\sqrt{2} - 5\sqrt{3}) = 9 \cdot 2 - 25 \cdot 3 = 18 - 75 = -57.

The rule in all three variants is the same: square the first term, square the second term, subtract. The radicals always die.

Where the shortcut matters most

Three recurring contexts:

Rationalising denominators. A denominator like \sqrt{5} + \sqrt{3} gets multiplied by its conjugate \sqrt{5} - \sqrt{3}, and the new denominator is (shortcut!) 5 - 3 = 2. Skipping the expansion saves a step every time.

Difference of squares in disguise. An expression like 7 - 3 = 4 appearing in an answer is often the result of this pattern, and recognising the structure backwards lets you factor 4 as (\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) when needed.

Simplifying complex rational expressions. Expressions with nested radicals in both numerator and denominator often have conjugate-pair factors hiding in them. Spotting a conjugate pair and collapsing it instantly is the fastest path through otherwise intimidating algebra.

A complete exam-style example

Simplify \dfrac{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})}{\sqrt{7} - \sqrt{3}}.

Step 1: spot the conjugate pair in the numerator. The product (\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) jumps to 7 - 3 = 4. Numerator becomes 4.

Step 2: the expression is now \dfrac{4}{\sqrt{7} - \sqrt{3}}. Rationalise by multiplying by the conjugate \sqrt{7} + \sqrt{3}:

\frac{4}{\sqrt{7} - \sqrt{3}} \cdot \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} = \frac{4(\sqrt{7} + \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} + \sqrt{3})}{4} = \sqrt{7} + \sqrt{3}.

Two uses of the same recognition reflex — once in the numerator, once in the rationalising step — and the whole thing is done in three lines. A student who insists on expanding term by term would take twice as long and probably make a sign error on the way.

The compressed reflex

See (\sqrt{a} \pm \sqrt{b})(\sqrt{a} \mp \sqrt{b}) → write a - b immediately. Never expand. The cross terms always cancel, the radicals always die, and recognising the pattern saves real time. Combined with the rationalising reflex and the denesting reflex from neighbouring articles, this is one of the three or four moves that define fluent surd algebra.