On nearly every Board paper and JEE Mains problem involving surds, there is a line where a fraction has a radical stuck in its denominator and the next step is to clean it up. The move is so mechanical that the fast students do not even pause — they see the radical, write "multiply top and bottom by the conjugate," and are done in the next line. The slow students stare at the denominator, try to simplify it with exponent rules, fail, and eventually ask for help.
The difference is a single pattern-recognition reflex. This article teaches the trigger, the conjugate that matches each denominator type, and the two-line execution.
The trigger
Scan for any of these shapes in the denominator of a fraction:
- A single radical: \sqrt{a}, \sqrt[3]{a}, or a multiple like 2\sqrt{3}.
- A sum or difference with one radical: a + \sqrt{b} or a - \sqrt{b}.
- A sum or difference of two radicals: \sqrt{a} + \sqrt{b} or \sqrt{a} - \sqrt{b}.
If the denominator matches any of these, the reflex is to multiply numerator and denominator by the conjugate of the denominator.
"Conjugate" means, roughly, "the thing that turns the denominator into a rational number when you multiply them." For each of the three shapes above, the conjugate has a specific form, and the product is guaranteed to collapse to a surd-free number.
The three conjugates
Shape 1: single radical \sqrt{a}. The conjugate is \sqrt{a} itself. Their product is (\sqrt{a})^2 = a.
Shape 2: binomial with one radical, p + q\sqrt{r}. The conjugate is p - q\sqrt{r} (flip the sign of the radical term). Their product is (p)^2 - (q\sqrt{r})^2 = p^2 - q^2 r, a rational number.
Shape 3: sum of two radicals, \sqrt{a} \pm \sqrt{b}. The conjugate is \sqrt{a} \mp \sqrt{b} (flip the sign between them). Their product is a - b.
Why all three conjugates collapse: each uses the difference-of-squares identity (x+y)(x-y) = x^2 - y^2. Any term that was a square root gets squared away, leaving only rational parts. That is the one algebraic fact powering the whole technique.
The two-line execution
Once you spot the trigger and identify the conjugate, the execution is always the same:
Line 1. Write the original fraction times \dfrac{\text{conjugate}}{\text{conjugate}}. This is multiplying by 1, so the value is unchanged.
Line 2. Simplify the denominator using the difference-of-squares collapse. The numerator usually just needs distribution.
Done.
Worked recognition: single-radical denominator
Problem: rationalise \dfrac{6}{\sqrt{3}}.
Spot the trigger: denominator is \sqrt{3}, a single radical. Conjugate: \sqrt{3}.
One line after the setup. Notice the numerator could be simplified at the end: 6/3 = 2.
Worked recognition: binomial denominator
Problem: rationalise \dfrac{1}{2 + \sqrt{3}}.
Spot the trigger: denominator is 2 + \sqrt{3}, shape p + q\sqrt{r} with p = 2, q = 1, r = 3. Conjugate: 2 - \sqrt{3}.
Why the denominator collapses: (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1. The (a+b)(a-b) = a^2 - b^2 pattern kills the radical cross-terms, leaving an integer.
Worked recognition: two-radical denominator
Problem: rationalise \dfrac{5}{\sqrt{7} - \sqrt{2}}.
Spot the trigger: denominator is \sqrt{7} - \sqrt{2}, shape \sqrt{a} - \sqrt{b}. Conjugate: \sqrt{7} + \sqrt{2}.
Three different denominator shapes, three conjugates, the same two-line execution.
Why the conjugate, specifically?
The conjugate is the unique factor that produces the difference-of-squares collapse. Any other multiplier would either leave a surd in the denominator or introduce new ones. Multiplying 2 + \sqrt{3} by 2 + \sqrt{3} (same sign, not flipped) would give 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} — still has a surd. Multiplying by 3 - \sqrt{3} would give 6 - 2\sqrt{3} + 3\sqrt{3} - 3 = 3 + \sqrt{3} — still has a surd. Only the sign-flipped version produces the clean cancellation.
So the reflex is not "multiply by something to kill the surd" — it is specifically "multiply by the conjugate," with the sign of the radical term flipped. Get that right and the difference-of-squares does the rest.
The one subtlety: when the numerator has a surd too
If the numerator itself contains a radical, distribute carefully on the top line. Example: rationalise \dfrac{\sqrt{3}}{\sqrt{5} + \sqrt{2}}.
Conjugate of denominator: \sqrt{5} - \sqrt{2}.
The top line uses \sqrt{a}\cdot\sqrt{b} = \sqrt{ab} to merge the products — but nothing else changes about the method. The denominator still collapses to 5 - 2 = 3 by difference of squares.
When to reach for this reflex
Every time a problem asks you to "simplify" or "rationalise," and the denominator contains any radical — single, binomial, or two-radical — this is the first move. It is the single most-used simplification in surd algebra, and it is the reason the standard form of answers in JEE solutions almost always has rational denominators.
Bonus: the same pattern works on complex-number denominators, where "conjugate" means flipping the sign of i. Multiplying a + bi by a - bi gives a^2 + b^2, clean and real. The surd trick and the complex trick are the same trick — difference-of-squares in two different costumes. See Complex Numbers: Conjugate and Modulus once you get there.
The compressed version
See a surd in a denominator → multiply top and bottom by the conjugate (flip the sign between the rational term and the radical, or between the two radicals) → difference-of-squares kills the radical in the denominator → simplify. Two lines of work, one reflex, zero surprises.
Related: Roots and Radicals · Rationalise the Denominator: Watch the Conjugate Flush the Irrational Upstairs · Conjugate Product: (√a + √b)(√a − √b) Jumps to a − b · Fractions and Decimals