You learned it in Class 8 as a harmless polynomial fact: (a - b)(a + b) = a^2 - b^2. The difference of squares. At the time, it looked like a tidy shortcut for multiplying out certain binomials — nothing dramatic.
That identity has a second life. In radical algebra, it is the engine that turns irrational denominators into rational ones. The same (a-b)(a+b) pattern, when a and b are square roots, has a superpower: the middle terms, which would each contain a radical, cancel out. Once you see why the cancellation happens — not just that it happens — the "conjugate trick" for rationalising stops feeling like magic. It becomes something you can reinvent from scratch whenever you need it.
Why the middle terms cancel
Take (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) and multiply it out term by term. Four products land:
Simplify each piece. \sqrt{a} \cdot \sqrt{a} = a and \sqrt{b} \cdot \sqrt{b} = b. The two middle pieces are +\sqrt{ab} and -\sqrt{ab} — opposite signs, identical magnitudes:
The middle pair — the two terms that carried a radical — cancelled exactly. What's left is a - b, rational if a and b are. Two radicals went in; one clean integer came out.
That's the whole mechanism. The cancellation happened because of the one-sign-flip between the two factors — the - in one binomial paired with the + in the other.
The habit
Whenever a binomial carries a radical — \sqrt{a} - \sqrt{b}, a - \sqrt{b}, \sqrt{a} + c, anything of that shape — and you want the product to come out rational, multiply by the conjugate: the same two terms with the middle operator flipped. Plus becomes minus, minus becomes plus.
The sign flip forces the cross-terms to carry opposite signs, so they cancel. You don't have to compute the cross-terms at all — skip straight to "a^2 - b^2" and write the answer.
Worked example 1 — (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})
Apply a^2 - b^2 with a = \sqrt{5} and b = \sqrt{2}:
Two radicals went in. One integer came out. If you had expanded the long way, the middle terms would have been +\sqrt{10} - \sqrt{10} = 0, but you don't need to check — the structure guarantees it.
Worked example 2 — (\sqrt{7} - 2)(\sqrt{7} + 2)
One side carries a radical, the other is an integer. Doesn't matter. Set a = \sqrt{7} and b = 2:
Again, the cross-terms were +2\sqrt{7} and -2\sqrt{7}, which summed to zero. The radical on one side and the integer on the other don't interfere with the pattern.
Worked example 3 — rationalising a denominator
Simplify \dfrac{1}{\sqrt{5} - \sqrt{3}}. The denominator is a binomial with two radicals. Multiply top and bottom by its conjugate \sqrt{5} + \sqrt{3}:
The denominator went from \sqrt{5} - \sqrt{3} (two radicals) to just 2 (a clean integer). The numerator picked up the conjugate, but that's fine — radicals upstairs are acceptable; it's radicals downstairs that block progress. See the rationalising habit for why.
Worked example 4 — when the numerator is also a binomial
Simplify \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}. Multiply top and bottom by 3 + \sqrt{2} (the conjugate of the denominator):
Numerator: (3 + \sqrt{2})(3 + \sqrt{2}) = (3 + \sqrt{2})^2. Use (a+b)^2 = a^2 + 2ab + b^2:
Denominator: (3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7.
Result: \dfrac{11 + 6\sqrt{2}}{7}.
The denominator cleared even though the numerator gained a radical. That's the whole point — the pipeline wants radicals up top, not down below. A denominator of 7 lets you add this to any other fraction, compare it to any other number, plug it into any further identity.
Why the cancellation is structural
When you expand (a - b)(a + b) by distributing, every factor on the left meets every factor on the right. Four products total:
- a \cdot a = a^2
- a \cdot b = ab
- (-b) \cdot a = -ab
- (-b) \cdot b = -b^2
Products (2) and (3) are the cross-terms. They are +ab and -ab. Their sum is zero. That's the cancellation.
Notice what controls the cancellation: the sign flip. If the second factor had been (a - b) instead of (a + b), you'd have (a-b)^2 — the cross-terms would be -ab and -ab, summing to -2ab, not zero. The + on one side and the - on the other is what forces the cross-terms to carry opposite signs.
This is why the trick is specifically about conjugates — the unique partner that flips exactly one sign. Any other pairing (same signs, both flipped, etc.) fails to produce the cancellation. The conjugate is not a convention. It is the single multiplier that kills the middle.
The cube-root analog
Cube roots have their own version, built on the identity (a - b)(a^2 + ab + b^2) = a^3 - b^3. Set a = \sqrt[3]{5} and b = \sqrt[3]{2}:
The three-term partner isn't exactly a conjugate — it's the thing that completes the cube. But the spirit is the same: the cross-terms, when you expand, arrange themselves to cancel, and only a^3 - b^3 survives. This is the tool for rationalising cube-root denominators. For n-th roots in general, the identity a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) provides the analogous multiplier.
When the conjugate trick doesn't work
The trick needs a binomial carrying a radical. If the expression isn't shaped like that, there is nothing for the middle to cancel.
- Monomial radical: \sqrt{5} \cdot \sqrt{3} = \sqrt{15}. Just multiply under the same root sign. No conjugate needed.
- Three-term radical: \sqrt{5} + \sqrt{2} + \sqrt{3}. A single conjugate flip won't collapse all the cross-terms — you'd need iterated conjugates or a cleverer grouping.
- Nested radicals: \sqrt{2 + \sqrt{3}}. Different machinery altogether — usually "denesting". The conjugate trick doesn't directly apply.
The trick is for a (\text{thing}) \pm (\text{thing}) shape where squaring the things makes them rational.
Why knowing the cancellation deepens understanding
Without understanding why the middle terms cancel, the conjugate trick is one more rule to remember: "when you see \sqrt{a} - \sqrt{b} in the denominator, multiply by \sqrt{a} + \sqrt{b}." But memorised rules slip, and they don't generalise — meet a cube-root denominator and the "rule" doesn't cover it.
Knowing the mechanism — that it's just (a-b)(a+b) = a^2 - b^2 applied to radical arguments, with cross-terms of opposite sign — lets you reinvent the trick for any situation. Cube roots? Use a^3 - b^3. Unusual binomial shape? Expand and find the multiplier that kills the cross-terms. The principle carries; the specific formula adapts.
Recognition drill
Try each of these in your head:
- (\sqrt{8} - \sqrt{3})(\sqrt{8} + \sqrt{3}) = 8 - 3 = 5.
- (\sqrt{11} + 1)(\sqrt{11} - 1) = 11 - 1 = 10.
- (2\sqrt{3} - 5)(2\sqrt{3} + 5) = (2\sqrt{3})^2 - 5^2 = 12 - 25 = -13. (The minus sign is fine — the identity gives a^2 - b^2, and sometimes b^2 > a^2.)
- \dfrac{1}{\sqrt{10} - 3} = \dfrac{1}{\sqrt{10} - 3} \cdot \dfrac{\sqrt{10} + 3}{\sqrt{10} + 3} = \dfrac{\sqrt{10} + 3}{10 - 9} = \sqrt{10} + 3.
Each one should take about five seconds. You don't multiply out; you just apply a^2 - b^2 with a and b as the two parts of the binomial.
The takeaway
The difference-of-squares identity isn't just a polynomial shortcut. In radical algebra, it is the engine that turns irrational denominators into rational ones — the cross-terms in the expansion have opposite signs and cancel. See a binomial with a radical; reach for the conjugate; write down a^2 - b^2. The middle terms disappear, and the problem opens up.