Algebra and the Manipulation of Surds for the ESAT

Understanding Surds and Conventions

In the context of the ESAT, a surd is an expression containing a root, usually a square root, which cannot be simplified into a rational expression. For example, $3 + 5\sqrt{2}$ is a surd because $\sqrt{2}$ is irrational and the expression cannot be simplified further. Conversely, $3 + 5\sqrt{64}$ is not a surd because $\sqrt{64} = 8$, allowing the expression to be simplified to $3 + 5(8) = 43$. Surds are the primary way mathematicians express numbers exactly, avoiding the imprecise and never-ending decimal expansions of irrational numbers like $\sqrt{2}$.

A vital convention used in the ESAT and TMUA is that the square root symbol $\sqrt{a}$ always refers to the positive root. Therefore, $\sqrt{64}$ is 8, not -8. If a question requires both the positive and negative roots, it will be written as $\pm\sqrt{64}$.

Simplifying Roots

You are expected to simplify expressions like $\sqrt{50}$ or $\sqrt{40}$ by identifying square factors. Consider the following examples:

1. $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$.
2. $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$. This can be further broken down into $2\sqrt{2 \times 5} = 2\sqrt{2}\sqrt{5}$.

Multiplying Expressions with Surds

When multiplying expressions containing surds, the most effective method is to treat the square roots like an algebraic variable, such as $x$, and then simplify the final constants. For example, to expand $(2 + 3\sqrt{5})^2$:

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

This follows the same pattern as $(2 + 3x)^2 = 4 + 12x + 9x^2$, where $x^2$ is replaced by $(\sqrt{5})^2 = 5$.

Factorising Expressions with Surds

While expanding $(2 + 3\sqrt{5})^2$ to get $49 + 12\sqrt{5}$ is direct, reversing the process to find the square root of $49 + 12\sqrt{5}$ requires a specific strategy. To find the exact value of $\sqrt{49 + 12\sqrt{5}}$, you must recognise the form $(a + b\sqrt{5})^2$.

Start with the middle term $12\sqrt{5}$. Since the middle term of $(a + b\sqrt{k})^2$ is $2ab\sqrt{k}$, we set $2ab\sqrt{5} = 12\sqrt{5}$, which means $ab = 6$. The constant part of the expression must satisfy $a^2 + k b^2 = 49$, which in this case is $a^2 + 5b^2 = 49$. By testing factor pairs of 6 for $a$ and $b$, we find that if $b = 3$ and $a = 2$, then $2^2 + 5(3^2) = 4 + 45 = 49$. Thus, the expression factorises to $(2 + 3\sqrt{5})^2$.

Be careful with subtractions. For $\sqrt{49 - 12\sqrt{5}}$, the answer is not $2 - 3\sqrt{5}$. Because $3\sqrt{5} = \sqrt{45}$, and $\sqrt{45} > 2$, the value $2 - 3\sqrt{5}$ is negative. Since the square root symbol must result in a positive value, the correct answer is $3\sqrt{5} - 2$.

Rationalising the Denominator

Rationalising involves moving surds from the denominator (bottom) to the numerator (top) so the denominator becomes a rational number. This is done by multiplying the fraction by a specific form of 1.

Simple Denominators

For an expression like $\frac{1}{\sqrt{5}}$, multiply by $\frac{\sqrt{5}}{\sqrt{5}}$:

$\frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

A general rule to remember is $\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$.

Compound Denominators

For more complex denominators, we use the difference of two squares: $(a + b)(a - b) = a^2 - b^2$. To rationalise $\frac{3}{2+\sqrt{5}}$, we multiply by $\frac{\sqrt{5}-2}{\sqrt{5}-2}$:

$\frac{3}{2+\sqrt{5}} \times \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{3\sqrt{5}-6}{5-4} = \frac{3\sqrt{5}-6}{1} = 3\sqrt{5}-6$.

Note that choosing $\sqrt{5}-2$ rather than $2-\sqrt{5}$ keeps the denominator positive, which simplifies the arithmetic.

Additional examples mentioned in the specification include:

1. $\frac{\sqrt{5}}{3+2\sqrt{5}}$: Multiply by $\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$ to get $\frac{3\sqrt{5}-10}{9-20} = \frac{3\sqrt{5}-10}{-11} = \frac{10-3\sqrt{5}}{11}$.

2. $\frac{3}{\sqrt{7}-2\sqrt{3}}$: Multiply by $\frac{\sqrt{7}+2\sqrt{3}}{\sqrt{7}+2\sqrt{3}}$ to get $\frac{3\sqrt{7}+6\sqrt{3}}{7-12} = \frac{3\sqrt{7}+6\sqrt{3}}{-5}$.

Advanced Extensions: Cube Roots

For denominators involving cube roots, you can apply the sum or difference of two cubes formulas: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

For example, to rationalise $\frac{x - 27}{\sqrt[3]{x} - 3}$, let $a = \sqrt[3]{x}$ and $b = 3$. The numerator $x-27$ is actually $a^3 - b^3$, which is $(a-b)(a^2+ab+b^2)$. The expression simplifies directly to $x^{2/3} + 3x^{1/3} + 9$.