Surds at Further Maths depth · Indices, logarithms & surds

In General Maths you rationalised denominators like $\sqrt3$ and $2 + \sqrt3$ (see surds). Further Maths questions put two surds on the bottom, ask for the answer in a set form such as $p + q\sqrt r$ , and add equations where the unknown is under a square root. The key tool is the same as before:

The difference of two squares(a + √b)(a − √b) = a² − b: the surds cancel

Two surds on the bottom

To clear a bottom such as $3\sqrt5 - 2\sqrt3$ , multiply the top and the bottom by its conjugate, the same two terms with the sign between them changed: $3\sqrt5 + 2\sqrt3$ . The bottom becomes a difference of two squares, $(3\sqrt5)^2 - (2\sqrt3)^2 = 45 - 12 = 33$ , with no surd left. The top needs all four products.

Rationalise with the conjugateStep through

3 + √23 − √2

The fractionstep 0 of 5

A surd with two terms on the bottom: 3 − √2. Multiplying by one surd would not clear it.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q3

Express $\dfrac{7 + 2\sqrt{10}}{9\sqrt5 + 5\sqrt2}$ in the form $p\sqrt5 + q\sqrt2$ .

Using the values of $p$ and $q$ in 3(a), find the value of $(2p - q)$ .

The conjugate
Change the sign between the terms: $9\sqrt5 - 5\sqrt2$ . Multiply the top and the bottom by it.
Think first. The bottom is 9√5 + 5√2. What is its conjugate?
The bottom
- It is a difference of two squares: $(9\sqrt5)^2 - (5\sqrt2)^2$ .
- Square both parts of each term: $81 \times 5 - 25 \times 2$ .
- So the bottom is ${405 - 50 = 355}$ .
Think first. (9√5)² − (5√2)² = ?
The top: four products
- ${7 \times 9\sqrt5 = 63\sqrt5}$
- ${7 \times (-5\sqrt2) = -35\sqrt2}$
- ${2\sqrt{10} \times 9\sqrt5 = 18\sqrt{50}}$
- ${2\sqrt{10} \times (-5\sqrt2) = -10\sqrt{20}}$
Simplify the surds
- ${\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2}$ , so ${18\sqrt{50} = 90\sqrt2}$ .
- ${\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt5}$ , so ${10\sqrt{20} = 20\sqrt5}$ .
- Collect the $\sqrt5$ terms: ${63\sqrt5 - 20\sqrt5 = 43\sqrt5}$ .
- Collect the $\sqrt2$ terms: ${90\sqrt2 - 35\sqrt2 = 55\sqrt2}$ .
- So the top is $43\sqrt5 + 55\sqrt2$ .
Think first. √50 and √20 each hide a square. Take it out.
(a) The form p√5 + q√2
- Divide each term by 355: $\dfrac{43}{355}\sqrt5 + \dfrac{55}{355}\sqrt2$ .
- Simplify the second fraction (divide by 5): ${\frac{55}{355} = \frac{11}{71}}$ .
- So ${p = \frac{43}{355}}$ and ${q = \frac{11}{71}}$ .
(b) 2p − q
- ${2p = \frac{86}{355}}$ , and ${q = \frac{11}{71} = \frac{55}{355}}$ .
- So ${2p - q = \dfrac{86 - 55}{355} = \dfrac{31}{355}}$ .

Adding and subtracting surd fractions

Two fractions whose bottoms are conjugates, such as $\sqrt5 - \sqrt2$ and $\sqrt5 + \sqrt2$ , share the common denominator $(\sqrt5 - \sqrt2)(\sqrt5 + \sqrt2) = 5 - 2 = 3$ . Put them over it in one step, instead of rationalising each one separately.

More: rationalising and surd fractions

Working with x = a + √b

When $x = a + \sqrt b$ and $a^2 - b = 1$ , the reciprocal is just the conjugate: $\frac1x = \frac{a - \sqrt b}{a^2 - b} = a - \sqrt b$ . So $x + \frac1x$ and $x - \frac1x$ come out neatly. For cubes, expand with $(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$ , simplifying each surd as you go.

More: expressions in a + √b

Equations with square roots

To solve an equation with the unknown under a square root:

get one square root on its own on one side;
square both sides (if a square root is still left, get it alone and square again);
solve the equation that’s left;
check every answer in the original equation. Squaring can add a false answer, because $(-2)^2$ and $2^2$ are both 4.

Square, then checkBoth roots of the quadratic must be tested in the first equation

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q4

If $\sqrt{3x + 1} - \sqrt{2x - 1} = 1$ , find the values of $x$ .

One root on its own
$\sqrt{3x + 1} = 1 + \sqrt{2x - 1}$ .
Think first. Move √(2x − 1) to the right-hand side.
Square both sides
- Square the left: $3x + 1$ .
- Square the right, keeping the middle term: ${1 + 2\sqrt{2x - 1} + (2x - 1)}$ .
- Tidy the right: ${2x + 2\sqrt{2x - 1}}$ .
- So ${3x + 1 = 2x + 2\sqrt{2x - 1}}$ .
- Take $2x$ from both sides, leaving the root on its own: ${x + 1 = 2\sqrt{2x - 1}}$ .
Think first. The right side is a bracket squared: (1 + √(2x − 1))². Expand it.
Square again
- Square both sides: ${x^2 + 2x + 1 = 4(2x - 1) = 8x - 4}$ .
- Bring everything to one side: ${x^2 - 6x + 5 = 0}$ .
- Factorise: ${(x - 1)(x - 5) = 0}$ , so ${x = 1}$ or ${x = 5}$ .
Check both
- ${x = 1}$ : ${\sqrt4 - \sqrt1 = 2 - 1 = 1}$ ✓.
- ${x = 5}$ : ${\sqrt{16} - \sqrt9 = 4 - 3 = 1}$ ✓.
- Both work, so ${x = 1}$ or ${x = 5}$ .
Think first. Put x = 1 and x = 5 into the first equation.

Surds inside other topics

Surd answers turn up in geometric progressions, trigonometry and the cosine rule. The rules are the same: simplify every surd, rationalise, and collect like terms.

More: surds in other topics

Your turn

WAEC 2016 · Paper 2 · Q3 (b)

(b)
Express $\dfrac{7\sqrt2 + 3\sqrt3}{4\sqrt2 - 2\sqrt3}$ in the form $p + q\sqrt r$ , where $p$ , $q$ and $r$ are rational numbers.

Worked solution (try it first)

(b)

Multiply the top and the bottom by the conjugate of the bottom, $4\sqrt2 + 2\sqrt3$ .
The bottom: $(4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20$ .
The top: $(7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18$
$= 74 + 26\sqrt6$ .
Divide each term by 20: $\dfrac{74}{20} + \dfrac{26}{20}\sqrt6 = \dfrac{37}{10} + \dfrac{13}{10}\sqrt6$ .
So $p = \frac{37}{10}$ , $q = \frac{13}{10}$ , $r = 6$ .

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