Simplifying surds using perfect-square factors, adding and subtracting like surds, multiplying them out, rationalising a denominator with the conjugate method, and solving equations that contain a surd — including the extraneous-root check that costs marks.
A surd is a root that cannot be simplified to a rational number, like √2 or √18. Grade 11 asks you to write them in simplest form and, crucially, to remove them from a denominator.
1Simplifying a surd
When the laws may be used. Over the real numbers, √ab = √a × √b needs a ≥ 0 and b ≥ 0, and the quotient law needs a ≥ 0 and b > 0 — a square root of a negative number is not a real number, and nothing may be divided by 0. Every surd in Grade 11 satisfies this, but say so when a question uses letters.
Look for the largest perfect square that divides into the number, then split the root.
- 1Find the largest perfect square factor of 72: 36.72 = 36 × 2.
- 2√72 = √36 × √2.Split the root over the product.
- 3= 6√2.√36 = 6 exactly.
Using a smaller factor still works but takes longer: √72 = √4 × √18 = 2√18, which must then be simplified again. Always hunt the largest square first.
2Adding and subtracting
Surds behave like like terms: you can only add or subtract them once the surd part matches.
- 1Simplify the second: √50 = √25 × √2 = 5√2.Now both terms contain √2.
- 23√2 + 5√2 = 8√2.Add the coefficients, keep the surd.
√a + √b is not √(a + b). Check it: √9 + √16 = 3 + 4 = 7, but √25 = 5.
3Multiplying
- 1√3 × √3 = 3.A root times itself gives the number.
- 2−5√3 + 2√3 = −3√3.Collect the surd terms.
- 32 × (−5) = −10.
- 4= 3 − 3√3 − 10 = −7 − 3√3.
4Rationalising the denominator
A final answer may not have a surd in the denominator. Multiply top and bottom by whatever clears it — this changes the form, not the value.
- 1Multiply top and bottom by √3.Multiplying by (√3)/(√3) = 1 leaves the value unchanged.
- 2(6√3)/3.√3 × √3 = 3.
- 3= 2√3.Simplify the fraction.
5The conjugate (two terms in the denominator)
(a + √b)(a − √b) = a2 − b
- 1Multiply top and bottom by the conjugate 1 − √2.A difference of squares kills the surd.
- 2Denominator: (1)² − (√2)² = 1 − 2 = −1.
- 3Numerator: 4(1 − √2) = 4 − 4√2.
- 4= (4 − 4√2)/−1 = −4 + 4√2.Divide through by −1.
Only the sign between the terms changes in a conjugate. It works because it creates a difference of two squares, and squaring a root removes it.
6Solving an equation that contains a surd
The ATP lists “solve simple equations involving surds” as a skill in its own right. The method is short — get the root alone, then square both sides — but it carries a trap that costs marks in every paper it appears in.
The condition matters because a square root is never negative. √x = −3 has no solution at all: no number you square gives a negative root. Squaring both sides would “give” x = 9, which is simply wrong.
- Step 1 — isolate. Get the surd on its own before you square anything.
- Step 2 — square both sides. Square the whole side, not each term.
- Step 3 — solve the equation that is left (often a quadratic).
- Step 4 — CHECK every answer in the ORIGINAL equation. This step is not optional and it is usually worth a mark.
- 13√x = 12Isolate the surd first — add 2 to both sides.
- 2√x = 4Divide by 3. The root is now alone.
- 3x = 16Square both sides.
- 4Check: 3 × 4 − 2 = 10 ✓Substitute back into the original equation.
7Why you must check: extraneous roots
Squaring is not reversible. Both 2 and −2 square to 4, so squaring an equation can create a solution the original equation never had. A value like that is called an extraneous root, and you find it only by substituting back.
- 1x + 2 = x2Square both sides. The surd is already alone.
- 2x2 − x − 2 = 0Write the quadratic in standard form.
- 3(x − 2)(x + 1) = 0, so x = 2 or x = −1Factorise and solve.
- 4Check x = 2: LHS = √4 = 2, RHS = 2 ✓This one works.
- 5Check x = −1: LHS = √1 = 1, RHS = −1 ✗A root is never negative, so this fails.
- 6∴ x = 2 onlyx = −1 is an extraneous root, created by the squaring.
If the right-hand side contains the variable, expect an extraneous root and leave room for the check. A quadratic that gives two answers where one is negative and the right-hand side must be positive is the standard exam setup.
Squaring term by term. (√x + 1)2 is not x + 1 — it is x + 2√x + 1. Square the whole side as one bracket, or isolate the surd first so you never have to.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Simplify √72
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6√2 - 2Simplify √50
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5√2 - 3Simplify √98
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7√2 - 4Simplify √12 + √27
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2√3 + 3√3 = 5√3 - 5Simplify 3√2 + √50
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8√2 - 6Simplify √5 × √20
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√100 = 10 - 7Expand (√3 + 2)(√3 − 5)
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−7 − 3√3 - 8Rationalise 6/(√3)
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2√3 - 9Rationalise 1/(√5)
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(√5)/5 - 10Rationalise 4/(1 + √2)
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−4 + 4√2 - 11Solve for x: √(x − 4) = 5
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x − 4 = 25 ✓; x = 29 ✓; check: √25 = 5 ✓ - 12Solve for x: 2√x + 1 = 9
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2√x = 8 ✓; √x = 4 ✓; x = 16 ✓ - 13Solve for x: √(x + 6) = x, and reject any invalid answer with a reason.
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x + 6 = x2 ✓; x2 − x − 6 = 0, so (x − 3)(x + 2) = 0 ✓; x = 3 or x = −2 ✓; x = −2 is rejected because √4 = 2, not −2, so x = 3 ✓ - 14Explain why √(x + 1) = −3 has no solution.
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A square root is never negative ✓; so no value of x can make the left side equal −3 ✓
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