AFCAT Numerical Ability · Squares, Cubes and Roots

Surds: Multiplying, Simplifying and Rationalising

Multiply and divide roots under one sign, take square and cube factors out of a root, square a two-term surd and undo it, and clear a root from a denominator.

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Why this matters

About a third of this page is HARD, mostly rationalising and the longer surd expressions. The rules are few: roots of one kind multiply under one sign, like surds add, and a conjugate clears a denominator. Marks go on adding unlike surds and on multiplying only the bottom of a fraction.

Concept 1 of 4: Multiplying and dividing roots

Roots of the same kind multiply and divide under one root sign. Choose the pairing so that what ends up under the sign is a perfect square or cube, and the root disappears.

Definition

$\sqrt a \times \sqrt b = \sqrt{ab}$ and $\dfrac{\sqrt a}{\sqrt b} = \sqrt{\dfrac ab}$ : the product rule and the quotient rule.
The same holds for cube roots: $\sqrt[3]{9} \times \sqrt[3]{3} = \sqrt[3]{27} = 3$ .
Look for a perfect power under the sign: $\sqrt 3 \times \sqrt{27} = \sqrt{81} = 9$ .
Only roots of the same kind combine like this: two square roots, or two cube roots.

Product and quotient rules

\sqrt[n]{a}\,\sqrt[n]{b} = \sqrt[n]{ab}, \quad \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac ab}

Worked example

Find

\sqrt{20} \times \sqrt 5

and

\dfrac{\sqrt[3]{81}}{\sqrt[3]{3}}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Squares, Cubes and Roots · Simplifying and Rationalising Surds

Simplify:

\sqrt{18} \times \sqrt{2}

Products combine, sums do not

\sqrt 3 + \sqrt{12}

is not

\sqrt{15}

. Only a product or a quotient goes under one root sign.

One kind of root at a time

\sqrt 5 \times \sqrt[3]{5}

is neither

\sqrt{25}

nor

\sqrt[3]{25}

. A square root and a cube root combine only as fractional powers, which come on the last page.

Drill 8 more on multiplying and dividing roots

Concept 2 of 4: Simplest form of a surd, and like surds

Take the largest square factor out of a square root, or the largest cube factor out of a cube root. Surds with the same root left over are like terms and add like

x

Definition

$\sqrt{a^2 b} = a\sqrt b$ : $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt 5$ .
$\sqrt[3]{a^3 b} = a\sqrt[3]{b}$ : $\sqrt[3]{40} = \sqrt[3]{8 \times 5} = 2\sqrt[3]{5}$ .
Like surds add and subtract through their numbers in front: $4\sqrt 5 - \sqrt 5 = 3\sqrt 5$ .
Unlike surds, such as $\sqrt 2$ and $\sqrt 3$ , do not combine.
If the value of one root is given, write every term as a multiple of that root first.

Taking out a square or cube factor

\sqrt{a^2 b} = a\sqrt b, \quad \sqrt[3]{a^3 b} = a\sqrt[3]{b}

Worked example

Simplify

\sqrt{98} + \sqrt{18} - \sqrt{72}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 2 · Squares, Cubes and Roots · Simplifying and Rationalising Surds

Simplify:

\sqrt{12} + \sqrt{27} - \sqrt{3}

Take out the largest square

\sqrt{200} = 2\sqrt{50}

is true but not finished. The largest square factor is 100, so

\sqrt{200} = 10\sqrt 2

Unlike surds stay apart

2\sqrt 5 + 3\sqrt 2

cannot be written as

5\sqrt 7

5\sqrt{10}

. It is already as simple as it gets.

Drill 8 more on simplest form of a surd, and like surds

Concept 3 of 4: Squaring a two-term surd, and its square root

(\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}

. Read backwards, it finds the square root of a number like

x + 2\sqrt y

: look for two numbers that add to

x

and multiply to

y

Definition

$(\sqrt a \pm \sqrt b)^2 = a + b \pm 2\sqrt{ab}$ , the square identities from the first page.
Backwards: $\sqrt{x \pm 2\sqrt y} = \sqrt a \pm \sqrt b$ , where $a + b = x$ , $ab = y$ and $a$ is the larger.
If the middle term has no 2 in front, make one: $16 + \sqrt{240} = 16 + 2\sqrt{60}$ . Then $10 + 6 = 16$ and $10 \times 6 = 60$ , so the root is $\sqrt{10} + \sqrt 6$ .

Square of a two-term surd

(\sqrt a \pm \sqrt b)^2 = a + b \pm 2\sqrt{ab}

Worked example

Find

(\sqrt{10} - \sqrt 3)^2

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Squares, Cubes and Roots · Simplifying and Rationalising Surds

What is one of the square roots of

9 - 2\sqrt{14}

Keep the cross term

(\sqrt{11} + \sqrt 5)^2

16 + 2\sqrt{55}

, not 16. The cross term

2\sqrt{ab}

is the part most often dropped.

Larger first with a minus

A square root of

13 - 2\sqrt{40}

\sqrt 8 - \sqrt 5

, that is

2\sqrt 2 - \sqrt 5

. Writing

\sqrt 5 - \sqrt 8

gives a negative number.

Drill 1 more on squaring a two-term surd, and its square root

Concept 4 of 4: Rationalising a denominator

Rationalising removes the root from the bottom of a fraction. Multiply top and bottom by whatever makes the bottom a whole number: the same root for a single square root, the conjugate for

\sqrt a \pm \sqrt b

Definition

$\dfrac{1}{\sqrt a} = \dfrac{\sqrt a}{a}$ .
A cube root needs two more of the same factor: $\dfrac{1}{\sqrt[3]{3}} = \dfrac{\sqrt[3]{9}}{3}$ , because $\sqrt[3]{3} \times \sqrt[3]{9} = 3$ .
The conjugate of $\sqrt a + \sqrt b$ is $\sqrt a - \sqrt b$ , and their product is $a - b$ . So $\dfrac{1}{\sqrt a + \sqrt b} = \dfrac{\sqrt a - \sqrt b}{a - b}$ .
For two numbers one apart the bottom becomes 1: $\dfrac{1}{\sqrt{n + 1} - \sqrt n} = \sqrt{n + 1} + \sqrt n$ . In a long sum of such terms, most of them cancel.

Rationalising with the conjugate

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

Worked example

Rationalise

\dfrac{4}{\sqrt 7 - \sqrt 3}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Squares, Cubes and Roots · Simplifying and Rationalising Surds

What is

\frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - \sqrt{4}}

equal to?

Multiply the top too

Multiplying only the bottom changes the value.

\dfrac{1}{\sqrt 6 - 2}

becomes

\dfrac{\sqrt 6 + 2}{2}

only when top and bottom are both multiplied by

\sqrt 6 + 2

A cube root needs two more

To clear

\sqrt[3]{3}

, multiply by

\sqrt[3]{9}

, not by

\sqrt[3]{3}

\sqrt[3]{3} \times \sqrt[3]{3} = \sqrt[3]{9}

is still a root.

Drill 7 more on rationalising a denominator

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 (4)

Multiplying and dividing roots
Product and quotient rules
$\sqrt[n]{a}\,\sqrt[n]{b} = \sqrt[n]{ab}, \quad \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac ab}$
Simplest form of a surd, and like surds
Taking out a square or cube factor
$\sqrt{a^2 b} = a\sqrt b, \quad \sqrt[3]{a^3 b} = a\sqrt[3]{b}$
Squaring a two-term surd, and its square root
Square of a two-term surd
$(\sqrt a \pm \sqrt b)^2 = a + b \pm 2\sqrt{ab}$
Rationalising a denominator
Rationalising with the conjugate
$\dfrac{1}{\sqrt a + \sqrt b} = \dfrac{\sqrt a - \sqrt b}{a - b}$

Watch out for (8)

Products combine, sums do not
→ Multiplying and dividing roots
One kind of root at a time
→ Multiplying and dividing roots
Take out the largest square
→ Simplest form of a surd, and like surds
Unlike surds stay apart
→ Simplest form of a surd, and like surds
Keep the cross term
→ Squaring a two-term surd, and its square root
Larger first with a minus
→ Squaring a two-term surd, and its square root
Multiply the top too
→ Rationalising a denominator
A cube root needs two more
→ Rationalising a denominator

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