Surds and Algebraic Fractions

Exact numbers and surds

A square root such as $\sqrt{49}=7$ is rational. By contrast, $\sqrt{2}$ cannot be written as an exact fraction of integers: its decimal continues without ending or recurring. Writing $\sqrt{2}$ preserves its exact value, so it is a surd. Replacing it prematurely with a rounded decimal loses accuracy.

The product rule for non-negative numbers is

$\sqrt{a}\,\sqrt{b}=\sqrt{ab}.$

In particular, $\sqrt a\times\sqrt a=a$. This lets us extract square . For example,

$\sqrt{72}=\sqrt{36\times2}=6\sqrt2.$

To put a surd in simplest form, find the largest square factor of the number under the root. A simplified surd has no square factor greater than $1$ remaining beneath the root.

Do not split a root across addition: in general, $\sqrt{a+b}\ne\sqrt a+\sqrt b$. The multiplication rule does not create a corresponding addition rule.

Collecting like surd terms

Once surds are simplified, collect only terms with exactly the same surd and variable part. The surd behaves like an algebraic factor:

$5\sqrt2+\sqrt{72}=5\sqrt2+6\sqrt2=11\sqrt2.$

Similarly,

$2a\sqrt5+7a\sqrt5=9a\sqrt5.$

However, $3\sqrt2+4\sqrt3$ cannot be collected because the surd parts differ. An integer and a surd are not like terms either, so is already collected.

Multiplying and expanding with surds

Use the ordinary rules of algebra, then simplify any products of roots. For example,

$3\sqrt2\times4\sqrt6=12\sqrt{12}=24\sqrt3.$

When expanding brackets, every term in one bracket multiplies every term in the other. For instance,

$$(2+\sqrt3)(4-\sqrt3) =8-2\sqrt3+4\sqrt3-3 =5+2\sqrt3.$$

The step $\sqrt3\times\sqrt3=3$ is what turns one product into an integer. Expansion with is therefore the same process as ordinary expansion; only the final simplification is different.

Factorising expressions containing surds

Factorising reverses expansion. First simplify the surds, then identify a factor shared by every term. For example,

$10a\sqrt3+\sqrt{48}ac=10a\sqrt3+4ac\sqrt3=2a\sqrt3(5+2c).$

Both the numerical factor $2$, the variable $a$ and the surd $\sqrt3$ are common. Expanding $2a\sqrt3(5+2c)$ gives the original expression, providing a quick check.

Rationalising a single-surd denominator

A fraction can be rewritten so that its denominator is rational. Multiplying its numerator and denominator by the same non-zero expression is equivalent to multiplying by $1$:

$$\frac{4}{\sqrt5} =\frac{4}{\sqrt5}\times\frac{\sqrt5}{\sqrt5} =\frac{4\sqrt5}{5}.$$

The value has not changed, but the denominator is now rational because $\sqrt5\times\sqrt5=5$.

Rationalising a two-term denominator

For a denominator such as $3+\sqrt5$, multiply by its conjugate, $3-\sqrt5$. The middle terms cancel through the difference of two squares:

$$(3+\sqrt5)(3-\sqrt5)=3^2-(\sqrt5)^2=9-5=4.$$

Therefore,

$$\frac{2}{3+\sqrt5} =\frac{2(3-\sqrt5)}{(3+\sqrt5)(3-\sqrt5)} =\frac{6-2\sqrt5}{4} =\frac{3-\sqrt5}{2}.$$

Use the opposite sign in the conjugate: $a+\sqrt b$ pairs with $a-\sqrt b$, and vice versa.

Algebraic fractions: factors and restrictions

Algebraic fractions are Higher-tier content. They follow the same rules as numerical fractions, but their denominators contain algebraic expressions. Every denominator must be non-zero, so begin by recording any excluded values.

For example, in

$\frac{x^2-16}{x^2+5x+4},$

the denominator factorises as $(x+1)(x+4)$, so $x\ne-1$ and $x\ne-4$.

Cancellation means dividing the numerator and denominator by a common factor. Factorising first gives

$$\frac{x^2-16}{x^2+5x+4} =\frac{(x-4)(x+4)}{(x+1)(x+4)} =\frac{x-4}{x+1},$$

with the original restrictions $x\ne-1,-4$. Although the factor $(x+4)$ disappears, $x=-4$ remains excluded because it made the original fraction undefined.

You cannot cancel the $x$ in $\frac{x+4}{x+1}$: neither numerator nor denominator has $x$ as a factor of the whole expression. Terms joined by $+$ or must not be cancelled.

Powers can be simplified using index laws when they are factors. Provided the original denominator is non-zero,

$\frac{32x^3y^2}{8xy}=4x^2y.$

Here the original expression requires $x\ne0$ and $y\ne0$.

Multiplying and dividing algebraic fractions

To multiply fractions, multiply numerators together and denominators together. It is often more efficient to factorise and cancel common factors before completing the multiplication:

$$\frac{x^2}{8}\times\frac{4}{x+3} =\frac{x^2}{2(x+3)}, \qquad x\ne-3.$$

To divide by a fraction, multiply by its reciprocal:

$$\frac{x}{x+2}\div\frac{3}{x-1} =\frac{x}{x+2}\times\frac{x-1}{3} =\frac{x(x-1)}{3(x+2)}.$$

The original expressions show that $x\ne-2$ and $x\ne1$. Also, a divisor cannot equal zero; here its numerator is $3$, so it is never zero where defined.

Adding and subtracting algebraic fractions

As with numerical fractions, addition and subtraction require a common denominator. For example,

$$\frac{2}{x}+\frac{3}{x+1} =\frac{2(x+1)+3x}{x(x+1)} =\frac{5x+2}{x(x+1)},$$

where $x\ne0,-1$.

If denominators share factors, use the lowest common denominator rather than multiplying unnecessarily. Expand and collect the numerator carefully, especially after a subtraction sign, then factorise again to see whether the result can be simplified.

Surds

Extract square factors: