Manipulating Surds for the TMUA

What is a Surd?

A surd is an expression containing a root, usually a square root, that cannot be simplified into a rational number. For example, $3 + 5\sqrt{2}$ is a surd because $\sqrt{2}$ is irrational and the expression cannot be reduced to a fraction. Conversely, $3 + 5\sqrt{64}$ is not a surd because it simplifies to $3 + 5(8) = 43$, which is a rational integer.

Surds allow mathematicians to express numbers exactly. Because the decimal expansion of an irrational number like $\sqrt{2}$ is non-terminating and non-recurring, writing it as a decimal involves a loss of precision. In the TMUA, we follow the standard convention that the square root symbol $\sqrt{a}$ refers specifically to the positive square root. Therefore, $\sqrt{64}$ is $8$, not $-8$. If both roots are required, we write $\pm\sqrt{64}$.

Simplifying Square Roots

You are expected to simplify square roots by identifying square factors within the radicand. For instance:

1. $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$.
2. $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} = 2\sqrt{2 \times 5} = 2\sqrt{2}\sqrt{5}$.

Multiplying Surd Expressions

When multiplying expressions containing surds, treat the square roots as you would an algebraic variable like $x$, then simplify the numerical terms at the final stage. Consider the expansion of $(2 + 3\sqrt{5})^2$:

$(2 + 3\sqrt{5})^2 = 2^2 + 2(2)(3\sqrt{5}) + (3\sqrt{5})^2$ $= 4 + 12\sqrt{5} + (9 \times 5)$ $= 4 + 12\sqrt{5} + 45$ $= 49 + 12\sqrt{5}$

This follows the same pattern as $(2 + 3x)^2 = 4 + 12x + 9x^2$, substituting $x = \sqrt{5}$ and noting that $x^2 = 5$.

Factorising Expressions with Surds

While expanding $(2 + 3\sqrt{5})^2$ is straightforward, reversing the process to find the square root of $49 + 12\sqrt{5}$ requires a systematic approach. To factorise $49 + 12\sqrt{5}$ into the form $(a + b\sqrt{5})^2$, we expand the general form:

$(a + b\sqrt{5})^2 = a^2 + 5b^2 + 2ab\sqrt{5}$

By matching the terms, we see that $2ab = 12$ (so $ab = 6$) and $a^2 + 5b^2 = 49$. By testing integer pairs for $ab = 6$ such as $(1,6), (2,3), (3,2)$, and $(6,1)$, we find that $a=2$ and $b=3$ satisfies the second equation since $2^2 + 5(3^2) = 4 + 45 = 49$. Thus, $\sqrt{49 + 12\sqrt{5}} = 2 + 3\sqrt{5}$.

Note on Negative Signs: When factorising $\sqrt{49 - 12\sqrt{5}}$, you might find $2 - 3\sqrt{5}$. However, because the square root symbol must represent a positive value, and $3\sqrt{5} = \sqrt{45} > 2$, the correct answer is the absolute value: $3\sqrt{5} - 2$.

Rationalising the Denominator

Rationalising the denominator involves transforming a fraction so that the denominator contains no surds. In simple cases like $\frac{1}{\sqrt{a}}$, we multiply by $\frac{\sqrt{a}}{\sqrt{a}}$ to get $\frac{\sqrt{a}}{a}$.

For more complex denominators, we use the difference of two squares identity: $(a + b)(a - b) = a^2 - b^2$. By multiplying the numerator and denominator by the conjugate of the denominator, we eliminate the root.

Worked Example 1: Rationalise $\frac{\sqrt{5}}{3 + 2\sqrt{5}}$.

Multiply by the conjugate $3 - 2\sqrt{5}$:

$\frac{\sqrt{5}(3 - 2\sqrt{5})}{(3 + 2\sqrt{5})(3 - 2\sqrt{5})} = \frac{3\sqrt{5} - 2(5)}{3^2 - (2\sqrt{5})^2} = \frac{3\sqrt{5} - 10}{9 - 20} = \frac{3\sqrt{5} - 10}{-11} = \frac{10 - 3\sqrt{5}}{11}$.

Worked Example 2: Rationalise $\frac{3}{\sqrt{7} - 2\sqrt{3}}$.

Multiply by the conjugate $\sqrt{7} + 2\sqrt{3}$:

$\frac{3(\sqrt{7} + 2\sqrt{3})}{(\sqrt{7} - 2\sqrt{3})(\sqrt{7} + 2\sqrt{3})} = \frac{3\sqrt{7} + 6\sqrt{3}}{7 - (4 \times 3)} = \frac{3\sqrt{7} + 6\sqrt{3}}{7 - 12} = \frac{3\sqrt{7} + 6\sqrt{3}}{-5} = -\frac{3\sqrt{7} + 6\sqrt{3}}{5}$.

Advanced Rationalisation with Cubic Identities

You can extend these principles to cube roots using the identities $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. For example, to rationalise $\frac{7}{\sqrt[3]{36} + \sqrt[3]{6} + 1}$, we let $a = \sqrt[3]{6}$ and $b = 1$. The denominator is $a^2 + a + 1$. Multiplying by $(a - 1)$ gives $a^3 - 1^3 = 6 - 1 = 5$. Thus:

$\frac{7}{\sqrt[3]{36} + \sqrt[3]{6} + 1} \times \frac{\sqrt[3]{6} - 1}{\sqrt[3]{6} - 1} = \frac{7(\sqrt[3]{6} - 1)}{5}$.