01 · Collect like roots
Simplify completely.
√147 − 2√12 + √27
Hint
Each root has a square factor and simplifies to a multiple of √3.
Worked solution
√147 = 7√3, √12 = 2√3
√27 = 3√3
7√3 − 4√3 + 3√3 = 6√3
Understand · explore · practise
Surds and rationalising denominators
Keep roots exact, simplify surd expressions and use conjugates to remove roots from denominators.
Before you startSquare numbers, fractions and expanding brackets
01 / Exact roots
Keep the length exact.
A surd is an irrational root left in exact form, such as √2 or √7. Decimal approximations are useful for estimating, but they lose the exact value.
Look for a square factor inside a square root. Its square root can come outside.
√98 = √(49 × 2) = 7√2
The picture explains the factor 7. Forty-nine small squares of area 2 form a 7-by-7 array. Each small side has length √2; each large side has length 7√2.
√(ab) = √a √b (a, b ≥ 0)
√(a/b) = √a / √b (a ≥ 0, b > 0)
These are product and quotient rules. There is no corresponding rule √(a + b) = √a + √b.
√(x²) = |x|. A square-root length is non-negative, even when x itself is negative.
A root is a lengthArea model
Area scales by 49; side length scales by √49 = 7.
02 / Combine & multiply
Simplify first. Then collect like roots.
Once each root is simplified, matching surds behave like matching variable terms: 2√3 + 5√3 = 7√3. Terms involving different roots generally stay separate.
√98 − 2√8 + √18
= 7√2 − 4√2 + 3√2
= 6√2
To multiply expressions with surds, distribute every term just as you would with x. Use (√a)² = a when a root is multiplied by itself.
A quotient can sometimes be simplified before any rationalisation:
√54/√6 = √(54/6) = 3
A useful counterexample: √(9 + 16) = 5, but √9 + √16 = 7. Splitting a root across addition changes the value.
Treat the root as one quantityWorked example
(3 + 2√5)(1 − √5)
Multiply each term in the first bracket by each term in the second.
= 3 − 3√5 + 2√5 − 2(√5)²
There are four products.
= 3 − √5 − 10
Collect the √5 terms; replace (√5)² with 5.
= −7 − √5
Keep this exact unless an approximation is requested.
03 / One root below
Multiply by a carefully chosen 1.
Rationalising a denominator means rewriting a fraction so that its denominator is rational. The value of the fraction must stay the same.
Multiply the numerator and denominator by the same non-zero expression. That multiplies the fraction by 1.
5/√7 = (5√7)/(√7 × √7)
= 5√7/7
The denominator is now 7. There is no reason to convert √7 to a decimal.
Always simplify numerical factors afterwards. If the denominator also contains an ordinary coefficient, it stays in the multiplication.
Simplify before you finishWorked example
6/(2√3) = 3/√3
Cancel the common numerical factor 2.
= (3√3)/(√3 × √3)
Multiply top and bottom by √3.
= 3√3/3 = √3
Cancel the remaining factor 3.
Check: (2√3) × √3 = 6
Multiplying the answer by the original denominator recovers the numerator.
04 / Two terms below
Choose the opposite sign.
The conjugate of a + √b is a − √b. Multiplying the two makes the mixed root terms cancel:
(a + √b)(a − √b) = a² − b
For rational a and non-negative rational b, the result is rational. The original denominator and the conjugate you use must be non-zero.
2/(5 + √6)
= 2(5 − √6)/[(5 + √6)(5 − √6)]
= (10 − 2√6)/19
The same idea works with two different roots because (√a + √b)(√a − √b) = a − b.
A numerator containing a root too
(√13 + 2)/(√13 − 2)
= (√13 + 2)²/(13 − 4)
= (17 + 4√13)/9
The numerator must be multiplied too. Squaring √13 + 2 creates a middle term of 4√13.
Opposite signs remove the root termsInspect the working
(5 + √6)(5 − √6)
= 25 − 5√6
+ 5√6 − 6
= 25 − 6 = 19
The middle terms are opposites. Their sum is zero.
Watch the opposite root terms cancel
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / More involved forms
Find a useful form for the denominator.
A squared denominator is still an expression you can expand. Once it has two terms, choose its conjugate.
Alternatively, if you already know 1/(3 − √2) = (3 + √2)/7, square both sides to obtain the result in the worked example.
Two different brackets in the denominator
1/[(3 + √2)(1 − √2)]
= 1/(1 − 2√2)
= (1 + 2√2)/(1 − 8)
= −(1 + 2√2)/7
Expand the denominator first. Then rationalise the two-term expression that remains.
An equation with a surd coefficient
Solve 4 + x√3 = 3x. Collect all x terms together:
4 = x(3 − √3)
x = 4/(3 − √3)
= 4(3 + √3)/(9 − 3)
= 2 + (2/3)√3
The denominator is non-zero, so the division is valid. Substitution into the original equation verifies the answer.
A squared denominatorWorked example
1/(3 − √2)²
Expand the whole square, including the middle term.
= 1/(11 − 6√2)
9 − 6√2 + 2 = 11 − 6√2.
= (11 + 6√2)/(121 − 72)
Multiply top and bottom by the conjugate.
= (11 + 6√2)/49
Equivalently, square (3 + √2)/7.
06 / Your turn
An exact answer should survive a check.
Work without rounding the roots. For a rationalised answer, multiply it by the original denominator: you should recover the original numerator.
02 · Check an expansion
A student writes (2 + √7)² = 11. What is missing, and what is the correct result?
Hint
Write the square as (2 + √7)(2 + √7) and list all four products.
Worked solution
4 + 2√7 + 2√7 + 7
= 11 + 4√7
The two mixed products were omitted. They add; they do not cancel.
03 · Use a conjugate
Rationalise and simplify.
3/(4 − √7)
Hint
Multiply the numerator and denominator by 4 + √7.
Worked solution
3(4 + √7)/(16 − 7)
= (12 + 3√7)/9
= (4 + √7)/3
Check: [(4 + √7)/3](4 − √7) = (16 − 7)/3 = 3.
04 · Two roots
Express this quotient with a rational denominator.
(√11 − √3)/(√11 + √3)
Hint
Use √11 − √3 as the conjugate. The numerator becomes a square.
Worked solution
(√11 − √3)²/(11 − 3)
= (14 − 2√33)/8
= (7 − √33)/4
05 · A squared denominator
Rationalise and simplify.
2/(4 + √3)²
Hint
You can square the conjugate or expand the denominator before rationalising.
Worked solution
2/(19 + 8√3)
= 2(19 − 8√3)/(361 − 192)
= (38 − 16√3)/169
06 · Use exact geometry
A square has side √18 cm. A designer wants a rectangle 3 cm long and less than 3 cm wide with the same perimeter. Is that possible? Calculate the required width exactly.
Hint
The square’s perimeter is 4√18. For a rectangle, perimeter = 2(length + width).
Worked solution
2(3 + w) = 4√18 = 12√2
3 + w = 6√2
w = 6√2 − 3 cm
Since √2 > 1, the required width is greater than 3 cm. It is positive, but it cannot meet the designer’s requirement of being less than 3 cm wide.
07 · Distances between markers
Markers along a straight path are placed at √1, √2, √3, …, √(N + 1) metres from a fixed start, where N is a positive integer. Find the distance from the first marker to the last. Then express the kth gap, √(k + 1) − √k, as a fraction with numerator 1 and explain why adding the gaps gives the same total.
Hint
Subtract the first position from the last. For a single gap, multiply by the conjugate divided by itself.
Worked solution
Total distance = √(N + 1) − 1 m
√(k + 1) − √k
= [(k + 1) − k]/[√(k + 1) + √k]
= 1/[√(k + 1) + √k]
Adding consecutive gaps cancels each intermediate marker position: arriving at a marker and departing from it contribute opposite terms. Only the final position minus the first position remains. This is called a telescoping sum.
07 / Recap
Change the form, preserve the value.
- Simplify a root: pull out square factors.
- Add or subtract: combine only matching simplified roots.
- Multiply brackets: include the mixed products.
- One root in the denominator: multiply top and bottom by that root.
- Two terms in the denominator: try the conjugate. Expand compound denominators when needed.
Keep every step exact. Round only if the question asks for a decimal approximation.
Section 1 of 7 · Exact roots