Indices, Radicals and Rational Expressions
Rationalising Denominators and Conjugates
Clearing a radical from a denominator, the conjugate pair and why it works, and the standard forms for one-term and two-term denominators.
What this page covers
- Rationalise a denominator containing a single radical
- Form the conjugate of a two-term surd expression and use it
- Explain why the conjugate removes the radical
- Recognise the same technique applied to complex numbers
The single-radical case
If the denominator is a lone radical, multiply top and bottom by that same radical. The denominator becomes the radicand, which is rational.
Multiplying by √x√x is multiplying by 1, so the value is untouched. Only the form changes — which is the fundamental identity for fractions, applied with an irrational multiplier.
Worked example — the source's preferred form
Converting to exponents made this immediate. The same result follows by multiplying top and bottom by √x: x3√xx = x2√x.
For an n-th root, multiply by whatever completes the power. To rationalise 13√x, multiply top and bottom by 3√x2, giving 3√x2x — because x1/3 needs x2/3 to reach x1.
The conjugate
A two-term denominator such as 2 - √5 cannot be cleared by multiplying by a single radical, because the rational term would then acquire one. The conjugate solves it.
The conjugate of a + √b is a - √b: the same terms with the connecting sign reversed. The source notes call 2 - √5 and 2 + √5 a conjugate pair.
The reason a conjugate works is the difference of two squares, and the source states the general identity before using it.
Both radicals disappear at once, because each is squared. The source checks it numerically:
| Expression | Conjugate | Product |
|---|---|---|
| 2 - √5 | 2 + √5 | 4 - 5 = -1 |
| √3 - √2 | √3 + √2 | 3 - 2 = 1 |
| √x + √a | √x - √a | x - a |
| 3 - 2i | 3 + 2i | 9 - 4i2 = 13 |
Worked examples
A rational and a surd — 32 - √5
The denominator came out negative, which is normal when the surd exceeds the rational term. Distributing the -1 over both terms of the numerator is the step where a sign is most often lost.
√5 ≈ 2.236, so the original is 3-0.236 ≈ -12.7, and -6 - 3(2.236) ≈ -12.7. Agreement.
Two surds — 5√3 - √2
The denominator became 1, so it vanished entirely. That is the ideal outcome and happens whenever the two radicands differ by 1.
Rationalising a numerator instead — √x - √ax - a
Sometimes the useful move is to clear the numerator. The source does exactly this, treating x - a in the denominator as the product of conjugates.
Recognising x - a as (√x + √a)(√x - √a) made the cancellation visible. The result is far better behaved: at x = a the original is 00, while the simplified form is 12√a, which is perfectly definite.
That last observation is why this manoeuvre matters well beyond tidiness. Clearing a radical from a numerator is the standard way of resolving an indeterminate 00 form, and it is the first technique met when derivatives of root functions are computed from first principles.
Choosing the multiplier
| Denominator shape | Multiply by | Denominator becomes |
|---|---|---|
| √b | √b | b |
| 3√b | 3√b2 | b |
| n√bm | n√bn-m | b |
| a + √b | a - √b | a2 - b |
| a - √b | a + √b | a2 - b |
| √a + √b | √a - √b | a - b |
| a + bi | a - bi | a2 + b2 |
The pattern throughout is the same: find the factor that turns the denominator into a difference of squares, or completes a fractional power to a whole one.
Multiply both numerator and denominator. Multiplying only the denominator changes the value of the expression rather than its form, which is a different and much worse kind of error than an untidy answer.
Common mistakes
| Mistake | Correct | Remedy |
|---|---|---|
| Multiplying only the denominator | Multiply top and bottom by the same quantity | The multiplier must equal 1 overall |
| Using 2 + √5 as the conjugate of 2 + √5 | The conjugate is 2 - √5 | The connecting sign reverses |
| 6 + 3√5-1 = -6 + 3√5 | -6 - 3√5 | The -1 multiplies both terms |
| Multiplying by 3√b for a cube-root denominator | Multiply by 3√b2 | b1/3 · b1/3 = b2/3, still irrational |
| Expanding the conjugate product term by term and slipping | (a + √b)(a - √b) = a2 - b | Use the identity directly; the cross terms always cancel |
| Rationalising when it hides a cancellation | Check whether clearing the numerator is more useful | √x - √ax - a is the classic case |
Frequently asked questions
Why bother rationalising at all?
Historically it made hand computation easier: dividing by a whole number is simpler than dividing by a decimal approximation to a surd. It remains a standard form, so answers are comparable, and it often reveals cancellation that the original form hides.
What exactly is a conjugate?
For a + √b the conjugate is a - √b — the same two terms with the sign between them reversed. Their product is a2 - b, which contains no radical, by the difference-of-two-squares identity.
Does the same idea work for complex numbers?
Yes, and it is the same identity. The conjugate of 3 - 2i is 3 + 2i, and their product is 9 - 4i2 = 9 + 4 = 13, a real number. Clearing i from a denominator and clearing a surd from one are the same manoeuvre.
What if the denominator has three terms?
Group it into two parts and conjugate in stages, which may take two rounds. It works but gets messy; in practice a three-term surd denominator is usually a sign that the expression should have been simplified earlier.
Source. Handwritten teaching notes, Week 2, page 10.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.