CDS Mathematics · Surds, Indices and Simplification

Square Roots of Surds

A surd a + 2√b is a perfect square (√m + √n)² when m + n = a and mn = b, which makes its square root and its reciprocal easy.

All Surds, Indices and Simplification notes CDS Mathematics strategy

Why this matters

Eleven PYQs. The same few numbers return again and again — 7 + 4√3 = (2 + √3)² appears in four papers — and each time the question is the square root, or the square root plus its reciprocal, which is then a whole number.

Concept 1 of 2: The square root of a + 2√b

(\sqrt m + \sqrt n)^2 = m + n + 2\sqrt{mn}

. So to take the square root of

a + 2\sqrt b

, find two numbers with sum

a

and product

b

. A coefficient other than

2

in front of the root must be rewritten first.

Definition

$\sqrt{a \pm 2\sqrt b} = \sqrt m \pm \sqrt n$ where $m + n = a$ , $mn = b$ , $m > n$ .
Rewrite: $4\sqrt{14} = 2\sqrt{56}$ , $6\sqrt7 = 2\sqrt{63}$ , so $16 + 6\sqrt7$ needs $m + n = 16$ , $mn = 63$ .
The square root is the positive one: for $a - 2\sqrt b$ write the larger root first.
A cube root keeps the sign: $\sqrt[3]{-0.008} = -0.2$ .

Square root of a surd

\sqrt{a + 2\sqrt b} = \sqrt m + \sqrt n, \quad m + n = a,\ mn = b

Worked example

Find

\sqrt{11 + 2\sqrt{30}}

and

\sqrt{9 - 4\sqrt5}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q76Moderate

Example 1 · Surds, Indices and Simplification · Square Roots of Surds

What is the square root of

16 + 6\sqrt{7}

The principal root is positive

\sqrt{9 - 4\sqrt5}

\sqrt5 - 2

, not

2 - \sqrt5

: both square to the same number, but only the first is positive. One CDS paper printed only the negative form among its options.

Drill 6 more on the square root of a + 2√b

Concept 2 of 2: A surd plus its reciprocal

2 + \sqrt3

and

2 - \sqrt3

multiply to

1

, so each is the other's reciprocal. Adding them cancels the surd and leaves a whole number.

Definition

$(p + \sqrt q)(p - \sqrt q) = p^2 - q$ . When this is $1$ , $\dfrac{1}{p + \sqrt q} = p - \sqrt q$ .
So if $\sqrt x = p + \sqrt q$ with $p^2 - q = 1$ : $\sqrt x + \dfrac{1}{\sqrt x} = 2p$ .
Useful squares: $7 + 4\sqrt3 = (2 + \sqrt3)^2$ , $97 + 56\sqrt3 = (2 + \sqrt3)^4$ , $3 + 2\sqrt2 = (1 + \sqrt2)^2$ , $11 + 2\sqrt{30} = (\sqrt6 + \sqrt5)^2$ .

Conjugates with product 1

(2 + \sqrt3)(2 - \sqrt3) = 1

Worked example

x = 3 + 2\sqrt2

, find

\sqrt x + \dfrac{1}{\sqrt x}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q92Moderate

Example 2 · Surds, Indices and Simplification · Square Roots of Surds

a = \sqrt{7 + 4\sqrt{3}}

, then what is the value of

a + \frac{1}{a}

Square root first, then the reciprocal

For

x = 7 + 4\sqrt3

x + \dfrac1x = 14

but

\sqrt x + \dfrac{1}{\sqrt x} = 4

. Check whether the question has

x

\sqrt x

before adding.

Drill 3 more on a surd plus its reciprocal

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)

The square root of a + 2√b
Square root of a surd
$\sqrt{a + 2\sqrt b} = \sqrt m + \sqrt n, \quad m + n = a,\ mn = b$
A surd plus its reciprocal
Conjugates with product 1
$(2 + \sqrt3)(2 - \sqrt3) = 1$

Watch out for (2)

The principal root is positive
→ The square root of a + 2√b
Square root first, then the reciprocal
→ A surd plus its reciprocal

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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