Simplifying and Rationalising Surds
This page covers two exam-heavy skills: simplifying surds by taking out square factors (with a reliable prime factorisation route for tougher numbers) and rationalising the denominator for both single-term and two-term denominators. Every worked example shows one move per line, exactly as examiners expect. When you are ready, scroll down to the practice rooms for randomly generated, auto-marked questions with instant feedback.
Core Ideas
Simplifying a Surd (take out a square factor)
Identify the largest square factor under the root, split the number into a square times a leftover, separate the roots, then bring the square root out. Pair this with the prime factorisation method below to guarantee you have not missed a larger square factor.
💡 Example A
\( \sqrt{20} \)
\( = \sqrt{4\times5} \)
\( = \sqrt{4}\,\sqrt{5} \)
\( = 2\sqrt{5} \)
What's happening?
4 is the largest square factor of 20, so split 20 into \(4\times5\).
Separate the roots, then replace \( \sqrt{4} \) with 2.
💡 Example B
\( \sqrt{50} \)
\( = \sqrt{25\times2} \)
\( = \sqrt{25}\,\sqrt{2} \)
\( = 5\sqrt{2} \)
What's happening?
Choose 25, the largest square factor; choosing a smaller one (such as no square at all) leaves the surd unsimplified.
Separate, then simplify \( \sqrt{25}=5 \).
Prime Factorisation Method (guaranteed route)
When the square factor is not obvious (for example \( \sqrt{147} \) or \( \sqrt{756} \)), break the number into primes, pair equal primes, move one from each pair outside the root, and leave any unpaired primes inside.
💡 Example A
\( \sqrt{147} \)
\( = \sqrt{3\times7\times7} \)
\( = 7\sqrt{3} \)
What's happening?
A quick factor tree gives \(147=3\times7^2\).
The pair of 7s sends one 7 outside the root; the unpaired 3 stays inside.
💡 Example B
\( \sqrt{756} \)
\( = \sqrt{2^2\times3^3\times7} \)
\( = 2\times3\,\sqrt{3\times7} \)
\( = 6\sqrt{21} \)
What's happening?
The factor tree gives \(756=2^2\times3^3\times7\).
The pair of 2s gives a 2 outside; one pair of 3s gives a 3 outside; the leftover 3 and the 7 stay inside as \( \sqrt{21} \).
Rationalising the Denominator
Goal: remove all surds from the denominator by multiplying by a "clever 1".
Single-term denominator \( \dfrac{k}{c\sqrt{m}} \)
Multiply by \( \dfrac{\sqrt{m}}{\sqrt{m}} \). If a coefficient sits with the surd, only the surd is targeted: \( \sqrt{m}\times\sqrt{m}=m \). Then simplify.
💡 Example A
\( \dfrac{2\sqrt{12}}{3\sqrt{3}} \)
\( = \dfrac{2\sqrt{12}\,\sqrt{3}}{3\,(\sqrt{3}\,\sqrt{3})} \)
\( = \dfrac{2\sqrt{36}}{3\times3} \)
\( = \dfrac{2\times6}{9} = \dfrac{12}{9} = \dfrac{4}{3} \)
What's happening?
Multiply top and bottom by \( \sqrt{3} \) only; the 3 in front is left alone.
On top, \( \sqrt{12}\,\sqrt{3}=\sqrt{36}=6 \). On the bottom, \( \sqrt{3}\,\sqrt{3}=3 \).
Finish by cancelling the fraction fully.
💡 Example B
\( \dfrac{x}{\sqrt{5}} \)
\( = \dfrac{x\sqrt{5}}{\sqrt{5}\,\sqrt{5}} \)
\( = \dfrac{x\sqrt{5}}{5} \)
What's happening?
Multiply top and bottom by the same surd, \( \sqrt{5} \).
The denominator becomes the whole number 5; check nothing cancels and leave the answer in exact form.
Two-term denominator \( a\pm\sqrt{b} \): use the conjugate
Multiply by the conjugate (same numbers, opposite sign): \( (a+\sqrt{b})(a-\sqrt{b})=a^2-b \). Expand the numerator and simplify.
💡 Example A
\( \dfrac{5}{4+\sqrt{7}} \)
\( = \dfrac{5\,(4-\sqrt{7})}{(4+\sqrt{7})(4-\sqrt{7})} \)
\( = \dfrac{20-5\sqrt{7}}{16-7} \)
\( = \dfrac{20-5\sqrt{7}}{9} \)
What's happening?
The conjugate of \(4+\sqrt{7}\) is \(4-\sqrt{7}\).
The bottom is a difference of two squares: \(4^2-(\sqrt{7})^2=16-7=9\). No surds survive in the denominator.
💡 Example B
\( \dfrac{4+\sqrt{5}}{2-\sqrt{5}} \)
\( = \dfrac{(4+\sqrt{5})(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} \)
\( = \dfrac{13+6\sqrt{5}}{4-5} \)
\( = \dfrac{13+6\sqrt{5}}{-1} = -13-6\sqrt{5} \)
What's happening?
Multiply by the conjugate \(2+\sqrt{5}\).
Expanding the top: \(8+4\sqrt{5}+2\sqrt{5}+5=13+6\sqrt{5}\).
The bottom is \(4-5=-1\), so the final tidy step divides everything by \(-1\).
🔑 Key Points
- Single term \(c\sqrt{m}\): multiply by \( \sqrt{m} \). Two terms \(a\pm\sqrt{b}\): multiply by the conjugate.
- Use \( \sqrt{ab}=\sqrt{a}\,\sqrt{b} \) only for \(a,b\ge 0\).
- Prime factorisation is the safety net for tougher numbers: factor, pair equal primes, extract one from each pair.
- One move per line with equals signs vertically aligned; keep exact surd form unless decimals are requested.
⚠️ Common Pitfalls
- Rationalising the whole \(2\sqrt{m}\) instead of just the \( \sqrt{m} \).
- Mixing up \( (a+b)^2 \) with \( (a+b)(a-b) \); conjugates use the difference of two squares.
- Missing the largest square factor; a quick factor tree prevents this.
- Two equals signs on one line; keep one move per line.