AFCAT Numerical Ability · Indices and Surds

Conjugates, Rationalising, and the Square Root of a Surd

Expand squares and conjugate products of surds, clear a surd from a denominator by multiplying by its conjugate, and find the square root of an expression like a + 2√b.

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Why this matters

This is the hardest page of the chapter: most of its questions are HARD. They are denominators such as 3 − √2 that must be cleared, sums of two such fractions, and square roots of expressions such as 3 + 2√2. All of them rest on one identity, (a + b)(a − b) = a² − b², and the marks go on a lost middle term or a wrong sign in the conjugate.

Concept 1 of 3: Squares and conjugate products of surds

3 + \sqrt 2

and

3 - \sqrt 2

are conjugates: the same two terms with the sign between them changed. Their product loses the root completely, because the cross terms cancel. A square keeps its middle term, and that middle term is where the root lives.

Definition

$(a + b)(a - b) = a^2 - b^2$ : with surds, $(\sqrt p + \sqrt q)(\sqrt p - \sqrt q) = p - q$ .
$(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ .
$(\sqrt p)^2 = p$ , so the squared terms come out rational and only $2ab$ carries a root.
The difference of the two squares, $(a + b)^2 - (a - b)^2$ , is $4ab$ ; their sum is $2(a^2 + b^2)$ .

Conjugate product

(\sqrt p + \sqrt q)(\sqrt p - \sqrt q) = p - q

Worked example

Expand

(3 + \sqrt 2)^2

and

(\sqrt 7 + \sqrt 2)(\sqrt 7 - \sqrt 2)

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Indices and Surds · Rationalising and Surd Identities

Simplify:

(2 + \sqrt{3})^2

A square keeps its middle term

(1 + \sqrt 3)^2 = 1 + 2\sqrt 3 + 3 = 4 + 2\sqrt 3

, not

1 + 3 = 4

. Squaring each term alone drops the

2ab

A conjugate product has no root left

(\sqrt 5 + 1)(\sqrt 5 - 1) = 5 - 1 = 4

. If a root is still in your answer, a cross term was not cancelled.

Drill 2 more on squares and conjugate products of surds

Concept 2 of 3: Rationalising a denominator

To rationalise is to clear the root out of the bottom of a fraction. Multiplying top and bottom by the same thing does not change the value. Choose that thing so the bottom becomes rational:

\sqrt a

for a single root, the conjugate for a sum or difference.

Definition

$\dfrac{k}{\sqrt a} = \dfrac{k\sqrt a}{a}$ : multiply top and bottom by $\sqrt a$ .
$\dfrac{k}{a + \sqrt b} = \dfrac{k(a - \sqrt b)}{a^2 - b}$ , and $\dfrac{k}{a - \sqrt b} = \dfrac{k(a + \sqrt b)}{a^2 - b}$ .
The conjugate changes only the sign between the terms: for $3 - \sqrt 7$ it is $3 + \sqrt 7$ .
A sum of two such fractions: rationalise each one, then add. Often the roots cancel.

Multiply by the conjugate

\dfrac{1}{a + \sqrt b} = \dfrac{a - \sqrt b}{a^2 - b}

Worked example

Rationalise

\dfrac{6}{\sqrt 3}

and

\dfrac{4}{\sqrt 5 - 1}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 2 · Indices and Surds · Rationalising and Surd Identities

Find:

\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} - 1}

Square the surd, not the number inside

(4 + \sqrt 3)(4 - \sqrt 3) = 16 - 3 = 13

, because

(\sqrt 3)^2 = 3

. Writing

16 - 9

squares the 3 a second time.

Multiply the top as well

Multiplying only the bottom by the conjugate changes the value of the fraction. Top and bottom both take the same factor.

Change only the middle sign

The conjugate of

3 - \sqrt 7

3 + \sqrt 7

. Changing the sign of the 3 instead gives

(3 - \sqrt 7)(-3 + \sqrt 7) = -16 + 6\sqrt 7

, and the root stays in the bottom.

Drill 5 more on rationalising a denominator

Concept 3 of 3: The square root of a + 2√b

Since

(\sqrt m + \sqrt n)^2 = m + n + 2\sqrt{mn}

, an expression

a + 2\sqrt b

is a perfect square whenever two numbers add to

a

and multiply to

b

. Find those two numbers and the square root is

\sqrt m + \sqrt n

Definition

First make the root term look like $2\sqrt b$ : $6\sqrt 2 = 2\sqrt{18}$ , so $b = 18$ .
Find $m$ and $n$ with $m + n = a$ and $mn = b$ .
$\sqrt{a + 2\sqrt b} = \sqrt m + \sqrt n$ .
$\sqrt{a - 2\sqrt b} = \sqrt m - \sqrt n$ with $m > n$ , so the answer is positive.
A perfect square of letters works the same way, but a square root is never negative: $\sqrt{x^2} = |x|$ .

Square root of a surd expression

\sqrt{a \pm 2\sqrt b} = \sqrt m \pm \sqrt n \quad (m + n = a,\ mn = b,\ m > n)

Worked example

Find

\sqrt{8 + 2\sqrt{15}}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Indices and Surds · Rationalising and Surd Identities

Find:

\sqrt{5 - 2\sqrt{6}}

Make it 2√b before you look for the pair

\sqrt{11 + 6\sqrt 2}

, the pair does not multiply to 2. Write

6\sqrt 2 = 2\sqrt{18}

: the pair is 9 and 2, and the root is

3 + \sqrt 2

With a minus sign, the larger root goes first

\sqrt{7 - 2\sqrt{10}} = \sqrt 5 - \sqrt 2

. The order

\sqrt 2 - \sqrt 5

is negative, and a square root never is.

The root of a square is a modulus

\sqrt{(x - 4)^2} = |x - 4|

, not

x - 4

. The two agree only when

x \ge 4

Drill 3 more on the square root of a + 2√b

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Squares and conjugate products of surds
Conjugate product
$(\sqrt p + \sqrt q)(\sqrt p - \sqrt q) = p - q$
Rationalising a denominator
Multiply by the conjugate
$\dfrac{1}{a + \sqrt b} = \dfrac{a - \sqrt b}{a^2 - b}$
The square root of a + 2√b
Square root of a surd expression
$\sqrt{a \pm 2\sqrt b} = \sqrt m \pm \sqrt n \quad (m + n = a,\ mn = b,\ m > n)$

Watch out for (8)

A square keeps its middle term
→ Squares and conjugate products of surds
A conjugate product has no root left
→ Squares and conjugate products of surds
Square the surd, not the number inside
→ Rationalising a denominator
Multiply the top as well
→ Rationalising a denominator
Change only the middle sign
→ Rationalising a denominator
Make it 2√b before you look for the pair
→ The square root of a + 2√b
With a minus sign, the larger root goes first
→ The square root of a + 2√b
The root of a square is a modulus
→ The square root of a + 2√b

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