Every solution that has a fraction like \dfrac{1}{2 + \sqrt{3}} goes through the same move: multiply top and bottom by 2 - \sqrt{3}, and suddenly the denominator is an integer. Your teacher calls 2 - \sqrt{3} the conjugate surd of 2 + \sqrt{3}. Where does that word come from, and why does the trick always work?

What a conjugate surd is

Given a binomial expression that contains a surd — an expression of the form a + b\sqrt{c} or \sqrt{a} + \sqrt{b} — the conjugate surd is the same expression with the sign between the two terms reversed.

Notice what does not change: the two terms themselves, and the coefficients. Only the sign between them flips.

The word "conjugate" comes from the Latin conjugatus, meaning "yoked together." The two surds a + b\sqrt{c} and a - b\sqrt{c} are yoked as a matched pair because their product has a special property — it contains no surd at all.

Why the product is always rational

Multiply a binomial surd by its conjugate. Take a + b\sqrt{c} and a - b\sqrt{c}:

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

The cross-terms +ab\sqrt{c} and -ab\sqrt{c} cancel exactly. The remaining two terms are a^2 (rational, since a is rational) and -b^2 c (rational, since b^2 and c are rational). So the product is purely rational — no surd survives.

This is the difference-of-squares identity (X + Y)(X - Y) = X^2 - Y^2 applied with X = a and Y = b\sqrt{c}. The identity is what forces the cross-terms to cancel and both squares to become rational.

Why the cross-terms cancel: when you expand (a + b\sqrt{c})(a - b\sqrt{c}) using FOIL, you get a \cdot a + a\cdot(-b\sqrt{c}) + b\sqrt{c}\cdot a + b\sqrt{c}\cdot(-b\sqrt{c}) = a^2 - ab\sqrt{c} + ab\sqrt{c} - b^2 c. The two middle terms are equal and opposite. That cancellation is the whole trick.

Why the denominator gets cleared

If you have a fraction \dfrac{N}{a + b\sqrt{c}} and multiply top and bottom by the conjugate a - b\sqrt{c}:

\frac{N}{a + b\sqrt{c}} \cdot \frac{a - b\sqrt{c}}{a - b\sqrt{c}} = \frac{N(a - b\sqrt{c})}{a^2 - b^2 c}.

The denominator becomes a^2 - b^2 c, a rational number. The value of the fraction has not changed — you multiplied by 1 — but the form has changed: the surd moved from the denominator to the numerator.

The only reason this works is that the conjugate is the unique partner that produces the difference-of-squares pattern. No other multiplier would eliminate the surd without creating a new one.

Worked check

Let us compute (2 + \sqrt{3})(2 - \sqrt{3}) directly:

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

The product is exactly 1. So if the original fraction is \dfrac{1}{2 + \sqrt{3}}, multiplying top and bottom by 2 - \sqrt{3} gives

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

The original fraction has been rewritten without any surd in the denominator, at no cost.

What conjugate to use for each shape

The rule "flip the sign between the two terms" covers every case. Here are the three you will meet:

In every case, the product is difference-of-squares and contains no surd.

<label>Coefficient $a$: <input data-param="a" type="range" min="1" max="9" step="1" value="2"></label> <label>Coefficient $b$: <input data-param="b" type="range" min="1" max="5" step="1" value="1"></label> <label>Radicand $c$: <input data-param="c" type="range" min="2" max="11" step="1" value="3"></label>
(a + b√c)(a − b√c) = a² − b²c =

Change a, b, c and watch the product: it is always an integer, confirming the surd terms have cancelled out.

What if you pick the wrong "conjugate"?

Suppose you tried to clear the surd by multiplying (2 + \sqrt{3}) by (2 + \sqrt{3}) itself (same sign, not flipped). You would get

(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3},

which still has \sqrt{3} in it. The denominator would have become worse, not better.

Or suppose you multiplied by (3 - \sqrt{3}) — a random binomial that is not the conjugate. Then

(2 + \sqrt{3})(3 - \sqrt{3}) = 6 - 2\sqrt{3} + 3\sqrt{3} - 3 = 3 + \sqrt{3},

again with a leftover surd.

Only the sign-flipped partner — the conjugate — hits the difference-of-squares identity and kills every surd term.

Rationalise $\dfrac{4}{\sqrt{7} + \sqrt{3}}$.

The denominator is \sqrt{7} + \sqrt{3}, a sum of two surds. 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}.

The denominator collapsed to the integer 4 because (\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3. The final answer is surd-free in the denominator and factored cleanly in the numerator.

Why this is one line of work: spot the shape, flip the sign, difference-of-squares collapses, cancel the 4. No expansion, no calculator, no guesswork.

Why only the surd's sign flips, not the whole expression

A common confusion: if the conjugate of 2 + \sqrt{3} is 2 - \sqrt{3}, why is the conjugate of 2 - \sqrt{3} not -2 + \sqrt{3} (negating both terms)? Because the name "conjugate" is about undoing the cross-term, not about negating everything.

The conjugate's job is to produce the difference-of-squares pattern. The pattern is (X + Y)(X - Y) = X^2 - Y^2. To hit it, you need the same X and Y, just with opposite signs between them. Negating both X and Y would instead give (-X - Y), whose product with the original is -(X+Y)^2 = -X^2 - 2XY - Y^2 — still has a surd cross-term, -2XY. That is not what you want.

Only one sign flip — the one between the two terms — gives the cross-term cancellation.

The bridge to complex numbers

The same word, "conjugate," reappears in complex numbers. The conjugate of a + bi is a - bi, and

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2,

a real number. The trick to dividing by a complex number in Class 11 is exactly the surd trick: multiply top and bottom by the conjugate, let difference-of-squares (with a sign twist because i^2 = -1) collapse the denominator.

So the "conjugate surd" technique you learn in Class 9 is the same mathematical tool as the "complex conjugate" you will meet in Class 11. Both are sign-flipped partners that let you multiply away an unwanted kind of number.

Related: Roots and Radicals · Surd in the Denominator? Multiply Top and Bottom by the Conjugate · Conjugate Product: (√a + √b)(√a − √b) Jumps to a − b · Why 'Rationalise the Denominator' If the Value Doesn't Change?