SPI Mathematics · Surds, Indices and Simplification

Surds and Rationalising

A surd is a root that cannot be simplified to a rational number; you simplify it, compare it, and clear it from a denominator with its conjugate.

All Surds, Indices and Simplification notes

Why this matters

Eight questions in the bank, two EASY and six MODERATE, and two are from the 2022 paper. They ask you to take square roots of decimals, compare roots, clear a root from a denominator, and find the square root of a number like 7 plus a root.

Concept 1 of 3: Simplifying and comparing roots

Pull perfect squares out from under a root to make it simpler. To compare roots of different kinds, such as a square root and a cube root, raise both to the same power until the roots disappear.

Definition

$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ , so $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ .
Only like surds add: $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ , but $\sqrt{2} + \sqrt{3}$ stays as it is.
The square root of a decimal has half as many decimal places: $\sqrt{2.25} = 1.5$ , $\sqrt{0.0004} = 0.02$ .
To compare $\sqrt[m]{a}$ and $\sqrt[n]{b}$ , raise both to the power LCM $(m, n)$ and compare the whole numbers you get.
A quick estimate also works: $\sqrt[3]{40}$ lies between 3 and 4, since $3^3 = 27$ and $4^3 = 64$ .

Roots as powers

\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad \sqrt[n]{a} = a^{\frac{1}{n}}

Worked example

Which is greater,

\sqrt{2}

\sqrt[3]{3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q37Easy

Example 1 · Surds, Indices and Simplification · Surds and Rationalising

Which number is the greatest among the given alternatives.

Splitting a root over a sum

\sqrt{9 + 16} = \sqrt{25} = 5

, not

3 + 4 = 7

. A root splits over a product, never over a sum.

Getting the decimal places wrong

\sqrt{0.0049} = 0.07

, not 0.7. The number under the root has four decimal places, so its root has two.

Adding unlike surds

\sqrt{2} + \sqrt{3}

is not

\sqrt{5}

: it is about

1.41 + 1.73 = 3.14

, while

\sqrt{5}

is about 2.24.

Drill 1 more on simplifying and comparing roots

Concept 2 of 3: Rationalising the denominator

A root in the denominator is cleared by multiplying top and bottom by its conjugate, the same two terms with the sign between them changed. The product of the pair is a difference of squares, so the root disappears.

Definition

The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$ , and $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ .
The conjugate is the simplest rationalising factor: it turns the surd into a rational number.
Multiply the top and the bottom by it, so the value does not change.
For $\frac{1}{\sqrt{a}}$ , multiply by $\frac{\sqrt{a}}{\sqrt{a}}$ : $\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$ .
If $x = a + \sqrt{b}$ and $a^2 - b = 1$ , then $\frac{1}{x} = a - \sqrt{b}$ , so $x + \frac{1}{x} = 2a$ . For sums of squares use $x^2 + y^2 = (x + y)^2 - 2xy$ .

Conjugate pair

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

Worked example

Write

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

in the form

a + b\sqrt{5}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Surds, Indices and Simplification · Surds and Rationalising

\frac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}

where a and b are rational numbers then

Multiplying only the denominator

Multiplying the bottom alone by the conjugate changes the value. Multiply the top by the same factor.

Losing a negative denominator

For

2 - \sqrt{7}

the product with its conjugate is

4 - 7 = -3

. Carry that minus sign into every term of the answer.

Using the wrong square identity

x^2 + y^2 = (x + y)^2 - 2xy

, but

x^2 + xy + y^2 = (x + y)^2 - xy

. Expand

(x + y)^2

and take away only what is extra.

Drill 3 more on rationalising the denominator

Concept 3 of 3: Square root of a surd and endless roots

A number like

5 + 2\sqrt{6}

is often a perfect square in disguise:

(\sqrt{3} + \sqrt{2})^2

. Find two numbers whose sum is the whole part and whose product is what sits under the root. An endless root repeats itself, so call it

x

and the repeat gives an equation.

Definition

To find $\sqrt{a + 2\sqrt{b}}$ , find $m$ and $n$ with $m + n = a$ and $mn = b$ . Then $\sqrt{a + 2\sqrt{b}} = \sqrt{m} + \sqrt{n}$ .
First make the root term look like $2\sqrt{b}$ : $\sqrt{28} = 2\sqrt{7}$ .
For $\sqrt{a - 2\sqrt{b}}$ the answer is $\sqrt{m} - \sqrt{n}$ , with the larger of $m$ and $n$ first so the result is positive.
For $x = \sqrt{k + \sqrt{k + \sqrt{k + \ldots}}}$ , the part under the first root is $k + x$ , so $x^2 = k + x$ . Keep the positive root.

Square root of a surd and an endless root

\sqrt{(m + n) + 2\sqrt{mn}} = \sqrt{m} + \sqrt{n} \qquad x = \sqrt{k + x} \Rightarrow x^2 - x - k = 0

Worked example

Find the positive square root of

5 + 2\sqrt{6}

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Surds, Indices and Simplification · Surds and Rationalising

The value of

\sqrt{6 + \sqrt{6 + \sqrt{6........}}}

Not making the root term 2 root b first

11 + \sqrt{120}

, write

\sqrt{120} = 2\sqrt{30}

before looking for m and n. Then

m + n = 11

mn = 30

, giving

\sqrt{6} + \sqrt{5}

Keeping the negative root

The quadratic for an endless root has one negative solution. A square root cannot be negative, so that solution is rejected.

Putting the smaller number first

\sqrt{4 - 2\sqrt{3}}

\sqrt{3} - 1

, not

1 - \sqrt{3}

, which is negative.

Drill 1 more on square root of a surd and endless roots

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)

Simplifying and comparing roots
Roots as powers
$\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad \sqrt[n]{a} = a^{\frac{1}{n}}$
Rationalising the denominator
Conjugate pair
$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b \qquad (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$
Square root of a surd and endless roots
Square root of a surd and an endless root
$\sqrt{(m + n) + 2\sqrt{mn}} = \sqrt{m} + \sqrt{n} \qquad x = \sqrt{k + x} \Rightarrow x^2 - x - k = 0$

Watch out for (9)

Splitting a root over a sum
→ Simplifying and comparing roots
Getting the decimal places wrong
→ Simplifying and comparing roots
Adding unlike surds
→ Simplifying and comparing roots
Multiplying only the denominator
→ Rationalising the denominator
Losing a negative denominator
→ Rationalising the denominator
Using the wrong square identity
→ Rationalising the denominator
Not making the root term 2 root b first
→ Square root of a surd and endless roots
Keeping the negative root
→ Square root of a surd and endless roots
Putting the smaller number first
→ Square root of a surd and endless roots

Test yourself on a real paper

Sit a past SPI paper, timed and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Sit an SPI past paper Drill all 8 questions