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Number · Surds

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Expanding brackets with surds

Why surds multiply out exactly like algebra — with √a·√a collapsing to a — how squares like (4 − √2)² keep their middle term, and why conjugate pairs multiply to a whole number.

Number · Surds

Expanding brackets with surds

Why surds multiply out exactly like algebra — with √a·√a collapsing to a — how squares like (4 − √2)² keep their middle term, and why conjugate pairs multiply to a whole number.

Why it works

Brackets with surds expand by exactly the rules you already know from algebra — treat 3\sqrt{3} the way you'd treat xx. The only new move is that a×a\sqrt{a} \times \sqrt{a} collapses to aa, so surd expansions simplify after expanding in a way algebra can't.

Single brackets. 3(4+3)=43+33=43+3\sqrt{3}(4 + \sqrt{3}) = 4\sqrt{3} + \sqrt{3}\sqrt{3} = 4\sqrt{3} + 3. Multiply through, collapse any aa\sqrt{a}\sqrt{a}, collect. When multiplying mixed terms, coefficients multiply with coefficients and roots with roots: 23×57=(2×5)3×7=10212\sqrt{3} \times 5\sqrt{7} = (2\times5)\sqrt{3 \times 7} = 10\sqrt{21}.

Double brackets: four terms, always. (2+5)(3−5)(2 + \sqrt{5})(3 - \sqrt{5}) expands (FOIL) to

6−25+35−55=6+5−5=1+5.6 - 2\sqrt{5} + 3\sqrt{5} - \sqrt{5}\sqrt{5} = 6 + \sqrt{5} - 5 = 1 + \sqrt{5}.

The rational bits collect together, the 5\sqrt{5} bits collect together, and the answer lands in the form a+bca + b\sqrt{c} — which is how exam questions usually ask for it.

Squaring keeps its middle term. (4−2)2(4 - \sqrt{2})^2 is (4−2)(4−2)(4 - \sqrt{2})(4 - \sqrt{2}), four terms like any double bracket:

16−42−42+2=18−82.16 - 4\sqrt{2} - 4\sqrt{2} + 2 = 18 - 8\sqrt{2}.

The classic error is (4−2)2=16+2=18(4 - \sqrt{2})^2 = 16 + 2 = 18 — squaring each term alone and losing the −82-8\sqrt{2}. A square of a sum is never just the sum of the squares; check with numbers: (1+2)2=9(1+2)^2 = 9, not 1+4=51 + 4 = 5.

Conjugate pairs multiply to a whole number. For (5+3)(5−3)(5 + \sqrt{3})(5 - \sqrt{3}) the outer and inner terms cancel:

25−53+53−3=25−3=22.25 - 5\sqrt{3} + 5\sqrt{3} - 3 = 25 - 3 = 22.

In general (a+b)(a−b)=a2−b(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b — the difference of two squares with the surd squared away. This "multiply by the twin with the opposite sign" trick is the engine behind rationalising denominators, and it's why exam questions love rectangles with sides (3+5)(3 + \sqrt{5}) and (3−5)(3 - \sqrt{5}): the area is exactly 9−5=49 - 5 = 4, no surd in sight.