CDS Mathematics · Surds, Indices and Simplification

Rationalising and Conjugate Pairs

Multiplying by the conjugate removes a surd from a denominator; a surd and its conjugate have a rational sum and product, and a chain of rationalised terms telescopes.

All Surds, Indices and Simplification notes CDS Mathematics strategy

Why this matters

Eleven PYQs. Most give x and y as conjugate fractions and ask for x² + y², x³ − y³ or similar — find x + y and xy first, then use an identity. The telescoping sums (1/(√n + √(n + 1)) added many times) appear almost every other year.

Concept 1 of 2: Conjugate pairs x and y

\dfrac{\sqrt a + \sqrt b}{\sqrt a - \sqrt b}

and its upside-down twin are reciprocals, so their product is

1

. Rationalise one to get a clean form, and the sum is rational too. Every symmetric expression then follows.

Definition

Rationalise: $\dfrac{1}{\sqrt a - \sqrt b} = \dfrac{\sqrt a + \sqrt b}{a - b}$ .
If $x = \dfrac{\sqrt a + \sqrt b}{\sqrt a - \sqrt b}$ and $y = \dfrac1x$ : $xy = 1$ and $x + y = \dfrac{2(a + b)}{a - b}$ .
Then $x^2 + y^2 = (x + y)^2 - 2$ , $x - y = \pm\sqrt{(x + y)^2 - 4}$ , $x^3 - y^3 = (x - y)\left((x + y)^2 - 1\right)$ .

Rationalising

\dfrac{\sqrt a + \sqrt b}{\sqrt a - \sqrt b} = \dfrac{(\sqrt a + \sqrt b)^2}{a - b}

Worked example

x = \dfrac{\sqrt3 + \sqrt2}{\sqrt3 - \sqrt2}

and

y = \dfrac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}

, find

x^2 + y^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (I) 2018 — Elementary Mathematics · Q18Moderate

Example 1 · Surds, Indices and Simplification · Surds and Rationalisation

What is the value of

\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} - \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}

Sign of a difference

\dfrac{\sqrt a - \sqrt b}{\sqrt a + \sqrt b}

is the smaller of the two twins, so 'small minus large' is negative. An option with the right size but the wrong sign is always printed.

Drill 6 more on conjugate pairs x and y

Concept 2 of 2: Telescoping sums

Rationalising

\dfrac{1}{\sqrt n + \sqrt{n + 1}}

gives

\sqrt{n + 1} - \sqrt n

. Add many such terms and each root cancels with the next, leaving only the last minus the first.

Definition

$\dfrac{1}{\sqrt n + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt n$ .
So $\displaystyle\sum_{n = p}^{q} \dfrac{1}{\sqrt n + \sqrt{n + 1}} = \sqrt{q + 1} - \sqrt p$ .
$\dfrac{1}{(n + 1)\sqrt n + n\sqrt{n + 1}} = \dfrac{1}{\sqrt n} - \dfrac{1}{\sqrt{n + 1}}$ , which telescopes the same way.

Telescoping term

\dfrac{1}{\sqrt n + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt n

Worked example

Find

\dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt5 + \sqrt6} + \cdots + \dfrac{1}{\sqrt{35} + \sqrt{36}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q12Moderate

Example 2 · Surds, Indices and Simplification · Surds and Rationalisation

What is the value of

\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{4}} + \ldots + \frac{1}{\sqrt{99}+\sqrt{100}}

Find the first and last terms exactly

A sum ending at

\dfrac{1}{\sqrt{195} + \sqrt{196}}

leaves

\sqrt{196}

, not

\sqrt{195}

. Write the first and last terms out before cancelling.

Drill 3 more on telescoping sums

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Conjugate pairs x and y
Rationalising
$\dfrac{\sqrt a + \sqrt b}{\sqrt a - \sqrt b} = \dfrac{(\sqrt a + \sqrt b)^2}{a - b}$
Telescoping sums
Telescoping term
$\dfrac{1}{\sqrt n + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt n$

Watch out for (2)

Sign of a difference
→ Conjugate pairs x and y
Find the first and last terms exactly
→ Telescoping sums

Test yourself on Surds, Indices and Simplification

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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