Open any JEE arithmetic problem and you will eventually meet an expression like \dfrac{1}{2 + \sqrt{3}} or (1 + \sqrt{5})(1 - \sqrt{5}) or a denominator that stubbornly contains \sqrt{7}. The moment you see the form p + q\sqrt{r} — where p, q, r are rational — a specific move should light up in your head: multiply by the conjugate p - q\sqrt{r}. That move turns a denominator with a surd in it into a clean rational denominator, and it is the single most reused simplification in the whole topic of surds.

This article teaches you the pattern, the algebra behind it, and when to reach for it on an exam.

The pattern

You see an expression of the form

p + q\sqrt{r} \qquad \text{or} \qquad \dfrac{\text{something}}{p + q\sqrt{r}}.

The conjugate is obtained by flipping the sign in front of the radical: p - q\sqrt{r}. Two facts make the conjugate magical.

Fact 1. The product of a binomial surd with its conjugate is rational.

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

Why: this is the difference-of-squares identity (a + b)(a - b) = a^2 - b^2 with a = p and b = q\sqrt{r}. The \sqrt{r} gets squared away to r, leaving no radicals on the right-hand side.

Fact 2. The sum of a binomial surd with its conjugate is twice the rational part.

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

This is also all-rational.

So the conjugate is a factor that, when multiplied in, erases the surd. That makes it the natural tool for cleaning surds out of denominators.

The canonical move: rationalising a denominator

Rationalise $\dfrac{1}{2 + \sqrt{3}}$

The denominator 2 + \sqrt{3} has a surd in it. You cannot leave it like that — the standard form of an answer has all irrationals in the numerator, none in the denominator.

Step 1. Multiply top and bottom by the conjugate 2 - \sqrt{3}.

\dfrac{1}{2 + \sqrt{3}} \cdot \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}.

Why: multiplying by \tfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = 1 doesn't change the value of the fraction — it just rewrites it in a cleaner form. The conjugate in the denominator is what triggers Fact 1.

Step 2. Apply the difference of squares.

(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1.

Step 3. Read off.

\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}.

Clean. No surd in the denominator. The whole radical got absorbed into the numerator.

That is the prototype. You will see dozens of variations.

When you see it on an exam, what to do

Reach for the conjugate any time one of these signals fires.

Signal 1. A surd in the denominator. \dfrac{3}{5 - \sqrt{2}}, \dfrac{\sqrt{5}}{\sqrt{7} + 1}, \dfrac{1}{\sqrt{3} - \sqrt{2}}. Rationalise immediately.

Signal 2. A product of two binomials, one of which looks like a conjugate. If the problem shows you (3 + \sqrt{2})(3 - \sqrt{2}), don't FOIL it out digit by digit — recognise the pattern and write 9 - 2 = 7 in one line.

Signal 3. "Find a and b such that \dfrac{\dots}{\dots} = a + b\sqrt{r}." This is just asking you to rationalise the left-hand side.

Signal 4. Simplifying \sqrt{a + b\sqrt{c}} into p + q\sqrt{c}. Slightly different technique (square both sides, match rational and irrational parts), but the conjugate is often hiding inside.

Why denominators shouldn't contain surds

Pedagogically: because answers in standard form have rational denominators, and graders deduct marks for "denominator contains a radical."

Practically: because rational denominators make arithmetic easier. If you want to add \dfrac{1}{\sqrt{2}} and \dfrac{1}{2}, you cannot just find a common denominator until you rationalise:

\dfrac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}.

Now \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} = \dfrac{\sqrt{2} + 1}{2}, which is final-form answer-style.

Fundamentally: because p + q\sqrt{r} is a single well-defined real number, and multiplying numerator and denominator by its conjugate is an invertible operation that moves that messiness upstairs, where a clean fraction expects it.

A cleaner example with two surds

Rationalise $\dfrac{1}{\sqrt{7} - \sqrt{3}}$

The denominator here has two radicals. The conjugate flips the sign between them: \sqrt{7} + \sqrt{3}.

Step 1. Multiply top and bottom.

\dfrac{1}{\sqrt{7} - \sqrt{3}} \cdot \dfrac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} = \dfrac{\sqrt{7} + \sqrt{3}}{(\sqrt{7})^2 - (\sqrt{3})^2} = \dfrac{\sqrt{7} + \sqrt{3}}{7 - 3} = \dfrac{\sqrt{7} + \sqrt{3}}{4}.

Done in one step. The denominator was a difference of surds; the conjugate is the sum; their product is the difference of squares.

The general form and a quick-reference table

For any positive rationals a, b and any positive integer r that is not a perfect square:

Rationalising denominators: conjugate tableA two-column reference table. The left column lists surd denominators and their conjugates. The right column gives the resulting rational denominator after multiplying by the conjugate. Four rows cover the standard JEE cases: a plus root r, a minus root r, root a plus root b, and root a minus root b. Denominator Conjugate Product (rational) a + √r a − √r a² − r a − √r a + √r a² − r √a + √b √a − √b a − b √a − √b √a + √b a − b always: flip the sign between the rational and irrational parts
The four standard cases. In every row, the conjugate flips the sign between the two terms, and the product is the difference of squares — which collapses to a rational number because the $\sqrt{\;}$ squares away.

Three-term surds — slightly heavier

What if the denominator has three terms, like 1 + \sqrt{2} + \sqrt{3}? There isn't a single "conjugate" that kills everything in one shot. You group: treat 1 + \sqrt{2} as the "rational part" and \sqrt{3} as the "irrational part," and multiply by (1 + \sqrt{2}) - \sqrt{3}.

\big((1 + \sqrt{2}) + \sqrt{3}\big)\big((1 + \sqrt{2}) - \sqrt{3}\big) = (1 + \sqrt{2})^2 - 3 = 1 + 2\sqrt{2} + 2 - 3 = 2\sqrt{2}.

One surd survives. Now rationalise again against \sqrt{2}: multiply top and bottom by \sqrt{2} and you're done.

Two rounds of conjugate-multiplication usually suffice for anything the exam throws at you.

The underlying intuition

A number of the form p + q\sqrt{r} lives in a little two-dimensional world of its own — one axis for the rational part, one for the \sqrt{r} part. The conjugate is the reflection across the rational axis: it flips the surd part, keeps the rational part.

Just as multiplying a complex number by its conjugate gives you a real number (|z|^2), multiplying a binomial surd by its conjugate gives you a rational number. Same idea, same algebra. If you are comfortable with complex numbers, the surd conjugate should feel like its older cousin.

Speed drill: what to do when you see p + q\sqrt{r}

The reflex you are building:

  1. Identify the rational part p and the "\sqrt{r} part" q\sqrt{r}.
  2. The conjugate is p - q\sqrt{r}.
  3. If there is a surd denominator, multiply top and bottom by the conjugate.
  4. The new denominator is p^2 - q^2 r — a plain rational.

Ten seconds of pattern recognition, two lines of algebra, done. Over a JEE paper this one move probably saves you two or three full minutes across different questions.

Related: Number Systems · Why Do We Need Irrational Numbers at All? · Drag the Decimal Point: Scientific Notation and Standard Form · Root of a Non-Perfect-Square Integer: Default to Irrational