Surds: meaning and recognising irrational roots

$\sqrt{4} = 2$. Easy — it's a rational number. But $\sqrt{2} \approx 1.41421\ldots$ — the decimal continues forever without repeating. It can't be written as a fraction. Numbers like this — irrational square roots of non-perfect-square integers — are called surds.

Surds appear all over WAEC: in geometry (lengths of diagonals), in algebra (solutions to quadratics with the quadratic formula), in trigonometry (exact values of sine and cosine). WAEC tests surd manipulation directly in nearly every paper.

This lesson defines surds, distinguishes them from rationals, and introduces the key notation. The next lesson covers simplification, operations, and rationalising denominators.

Learning objectives

By the end of this lesson you should be able to:

• define a surd and distinguish surds from rational numbers
• recognise square roots, cube roots and higher roots that are surds
• use the product and quotient properties of surds
• connect surds with fractional indices

Recognising surds

Square roots

The square roots of perfect squares are NOT surds (they're integers): $\sqrt{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \ldots}$.

The square roots of all other positive integers ARE surds: $\sqrt{2, 3, 5, 6, 7, 8, 10, 11, 12, \ldots}$.

Higher roots

For cube roots: $\sqrt[3]{a}$ is a surd if $a$ is not a perfect cube.

Perfect cubes: $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000$.

So $\sqrt[3]{8} = 2$ (not a surd), $\sqrt[3]{27} = 3$ (not), but $\sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}$ are all surds.

Why surds matter

Surds can't be represented exactly by a terminating or repeating decimal. They must be expressed either:

• As the radical itself (e.g., $\sqrt{2}$).
• As a numerical approximation (e.g., $\sqrt{2} \approx 1.414$).

WAEC typically asks for the exact answer — in surd form — rather than a decimal approximation.

Figure 1. √2 is the diagonal of a unit square: an exact length that cannot be written as a fraction. WASSCE Tutor diagram.

Common surds and their approximate values

Knowing the approximate value of common surds helps with estimation:

Memorising $\sqrt{2}, \sqrt{3}, \sqrt{5}$ to two decimal places is sufficient for most WAEC questions.

Notation

• $\sqrt{a}$ or $a^{1/2}$ — the square root of $a$.
• $\sqrt[n]{a}$ or $a^{1/n}$ — the $n$-th root of $a$.
• $\sqrt[3]{a}$ — cube root.
• The expression inside the radical sign is called the radicand.
• A surd of the form $\sqrt[n]{a}$ where $a$ is positive and not a perfect $n$-th power is a pure surd.
• A surd combined with rational numbers, like $3\sqrt{2}$ or $2 + \sqrt{5}$, is a mixed expression.

Connection to indices

From the Indices topic:

$$\sqrt[n]{a} = a^{1/n}$$

So $\sqrt{2} = 2^{1/2}$, $\sqrt[3]{5} = 5^{1/3}$, $\sqrt[4]{7} = 7^{1/4}$.

This lets us use the laws of indices to manipulate surds.

Key properties of surds

Product property

$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$

This works for any $a, b \geq 0$.

Examples:

• $\sqrt{2} \times \sqrt{3} = \sqrt{6}$.
• $\sqrt{5} \times \sqrt{7} = \sqrt{35}$.
• $\sqrt{6} \times \sqrt{2} = \sqrt{12}$.
• $\sqrt{a} \times \sqrt{a} = \sqrt{a^2} = a$. (Squaring undoes square root.)

Quotient property

Examples:

• $\dfrac{\sqrt{8}}{\sqrt{2}} = \sqrt{\dfrac{8}{2}} = \sqrt{4} = 2$.
• $\dfrac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2$.
• $\dfrac{\sqrt{15}}{\sqrt{5}} = \sqrt{3}$.

Power property

$$(\sqrt{a})^n = a^{n/2}$$

For $n = 2$: $(\sqrt{a})^2 = a$. This is the inverse of taking the square root.

Negation of square root

The square root function returns the non-negative root by convention. So $\sqrt{4} = 2$, not $-2$. The negative root is $-\sqrt{4} = -2$.

Note: you can never take the square root of a negative number in the real number system. $\sqrt{-4}$ is not a real number. (It's a complex number, $2i$, beyond WAEC scope.)

Examples — distinguishing surds and rationals

Example 1

Which of the following are surds, and which are rationals?

(a) $\sqrt{16}$ — $\sqrt{16} = 4$, rational (not a surd). (b) $\sqrt{20}$ — $20$ is not a perfect square. Surd. (c) $\sqrt{49}$ — $\sqrt{49} = 7$, rational. (d) $\sqrt[3]{27}$ — $\sqrt[3]{27} = 3$, rational. (e) $\sqrt[3]{10}$ — $10$ is not a perfect cube. Surd. (f) $\sqrt{0.04}$ — $\sqrt{0.04} = 0.2$, rational. (g) $\dfrac{3}{\sqrt{2}}$ — contains a surd, the expression is irrational. (We would need to rationalise it: see the next lesson.) (h) $\sqrt{2} + 3$ — contains a surd, irrational.

Approximating surds

For problems where exact form isn't required, approximate by knowing nearby perfect squares.

Example 2

Estimate $\sqrt{50}$.

Perfect squares near $50$: $49$ ($7^2$) and $64$ ($8^2$). So $\sqrt{49} = 7, \sqrt{64} = 8$. Therefore $\sqrt{50}$ is between $7$ and $8$, closer to $7$.

More precisely: $7^2 = 49$, $7.1^2 = 50.41$. So $\sqrt{50} \approx 7.07$.

Example 3

Estimate $\sqrt{75}$.

$8^2 = 64, 9^2 = 81$. So between $8$ and $9$, closer to $9$. $8.5^2 = 72.25$, $8.7^2 = 75.69$. So $\sqrt{75} \approx 8.66$.

When the radical is in algebraic form

Surds also appear with variables:

• $\sqrt{x^2} = |x|$ (absolute value, since the square root is non-negative).
• $\sqrt{x^2 y^2} = |xy|$ if $x, y$ have known signs; or $|x| \cdot |y|$ in general.
• $\sqrt{4x} = 2\sqrt{x}$ (if $x \geq 0$).

WAEC usually works with positive variables, so we often drop the absolute value: $\sqrt{x^2} = x$ (assuming $x \geq 0$).

WASSCE examination focus

Practice problems

1. Which of the following are surds? (a) $\sqrt{36}$ (b) $\sqrt{50}$ (c) $\sqrt[3]{64}$ (d) $\sqrt[3]{20}$ (e) $\sqrt{1/4}$ (f) $\sqrt{0.16}$ (g) $\sqrt{17}$

2. Compute $\sqrt{12} \times \sqrt{3}$.

3. Compute $\sqrt{50} \div \sqrt{2}$.

4. Compute $(\sqrt{6})^2$.

5. Compute $\sqrt{9} + \sqrt{16}$.

6. Estimate $\sqrt{30}$ (state between which two integers).

7. Compute $\sqrt{2} \times \sqrt{8}$.

8. Compute $\sqrt{45} \div \sqrt{5}$.

9. Is $\sqrt{-4}$ defined in the real numbers?

10. True or False: $\sqrt{4 + 9} = \sqrt{4} + \sqrt{9}$.

Worked solutions

1. (b) $\sqrt{50}$, (d) $\sqrt[3]{20}$, (g) $\sqrt{17}$. The others reduce to rationals.

2. $\sqrt{12 \times 3} = \sqrt{36} = 6$.

3. $\sqrt{50/2} = \sqrt{25} = 5$.

4. $6$.

5. $3 + 4 = 7$.

6. $25 < 30 < 36$, so $5 < \sqrt{30} < 6$.

7. $\sqrt{16} = 4$.

8. $\sqrt{9} = 3$.

9. No, $\sqrt{-4}$ is undefined in real numbers.

10. False. $\sqrt{4 + 9} = \sqrt{13} \neq 7$.

What's next

In the next lesson we cover simplifying surds, adding, multiplying and rationalising denominators — extracting any perfect-square factors from the radicand. $\sqrt{50} = 5\sqrt{2}$, $\sqrt{75} = 5\sqrt{3}$. This is the most common surd-related WAEC operation.