2.2 Surds and rationalising the denominator — Learn

Prerequisites: the index laws from 2.1; square numbers and their factors; expanding brackets; the difference of two squares.

By the end of this section you can: simplify a surd; add, subtract and multiply surds; remove a surd from the denominator of a fraction, including a denominator of the form $a + \sqrt{b}$; and explain why a simplified surd is an exact answer.

1. What a surd is

A surd is a root that cannot be written as an exact fraction. The number $\sqrt{2}$ is a surd: it is about $1.414$, but no fraction equals it exactly. Leaving an answer as $\sqrt{2}$ keeps it exact, while $1.414$ is only an approximation.

Two laws let you move factors in and out of a root:

The first law in reverse is how a surd is simplified. Look for a square factor inside the root.

Example 1. Simplify $\sqrt{50}$.

Write $50$ as $25 \times 2$, and $25$ is a square:

Now take the square root of $25$:

The number $5\sqrt{2}$ is the simplest exact form.

2. Adding and subtracting surds

Surds behave like algebraic terms. You can add $3x$ and $5x$ because they are like terms; you can add $3\sqrt{2}$ and $5\sqrt{2}$ for the same reason. Surds with different numbers under the root are not like terms, but they may simplify to like terms.

Example 2. Simplify $\sqrt{12} + \sqrt{75}$.

Simplify each surd first:

and

Now the two surds are like terms, so add them:

3. Multiplying surds

Multiply the numbers inside the roots. A product of two surds often contains a square factor, so simplify again at the end.

Example 3. Work out $\sqrt{8} \times \sqrt{18}$.

Multiply inside the roots:

Then take the square root:

4. Rationalising the denominator

A fraction is in its simplest form when its denominator contains no surd. To remove a single surd from the denominator, multiply the top and the bottom by that surd.

Example 4. Write $\dfrac{6}{\sqrt{3}}$ with a whole-number denominator.

Multiply the top and the bottom by $\sqrt{3}$:

The denominator becomes $\sqrt{3} \times \sqrt{3} = 3$:

When the denominator is $a + \sqrt{b}$, multiply by its conjugate $a - \sqrt{b}$. The product is the difference of two squares, and the surd disappears.

Example 5. Write $\dfrac{1}{3 + \sqrt{2}}$ with a rational denominator.

Multiply the top and the bottom by the conjugate $3 - \sqrt{2}$:

The denominator is a difference of two squares:

So

Watch out. The trap is applying a product law to a sum.

$\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$. The laws $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ hold for a product and a quotient. There is no rule for a sum.

Re-check: after simplifying, square your answer, or substitute a decimal approximation into both the original and the answer. If the two decimals agree, the simplification is sound.

Progress check

Return to the chapter problem. The arch is modelled by $y = 4 - \frac{x^2}{4}$, and its edge meets the road at $x = 4$.

1. Work out the height of the arch at $x = 2$ and at $x = 3$.
2. Suppose a second model divides the height by $\sqrt{3}$. Write the new height at $x = 0$.
3. Explain why $\sqrt{2} + \sqrt{8}$ is $3\sqrt{2}$, not $\sqrt{10}$.

You should now be able to

• simplify a surd by removing its largest square factor;
• add and subtract surds that are like terms;
• multiply surds and simplify the result;
• rationalise a denominator of the form $\sqrt{a}$ or $a \pm \sqrt{b}$;
• explain why a simplified surd is exact, while a decimal is not.