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Surds & Indices

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CAT 2025 Lesson : Surds & Indices - Comparison of Surds

1.2 Comparison of surds

To compare two surds and find out which one is greater/smaller,

(1) Subtract/add the same integer to both surds to make them comparable.
(2) Square the surds if the numbers are close after step 1.
(3) To compare square roots, find their integral ranges.

Example 2

a = 2 + \sqrt{8}

b = 4 + \sqrt{3}

and

c = 5 - \sqrt{2}

. Arrange

a

b

and

c

in descending order.

Solution

We can compare two surds at a time.

a - 2 = \sqrt{8}

and

b - 2 = 2 + \sqrt{3}

2 < \sqrt{8} < 3

and

1 < \sqrt{3} < 2

\therefore (a - 2)

is between

2

and

3

(b - 2)

is between

3

and

4

\therefore b > a

b - 4 = \sqrt{3}

, which is between

1

and

2

c - 4 = 1 - \sqrt{2}

, which is between

-1

and

0

\therefore b > c

a - 2 = \sqrt{8}

, which is between

2

and

3

c - 2 = 3 - \sqrt{2}

, which is between

1

and

2

\therefore a > c

Answer:

b > a > c

Alternatively

\sqrt{8}

, which is between

2

and

3

, can be written as

2

.##, where ## are some digits after the decimal point.

Applying this to

a

b

and

c

, we get

a = 2 + \sqrt{8} = 2 + 2

.## =

4

.##

b = 4 + \sqrt{3} = 4 + 1

.## =

5

.##

c = 5 - \sqrt{2} = 5 - 1

.## =

3

.##

\therefore b > a > c

Example 3

a = \sqrt{7} + \sqrt{6}

and

b = \sqrt{3} + \sqrt{11}

. Which of

a

and

b

is greater?

Solution

a^{2} = (\sqrt{7} + \sqrt{6})^{2} = 13 + 2 \sqrt{42}

b^{2} = (\sqrt{3} + \sqrt{11})^{2} = 14 + 2 \sqrt{33}

a^{2} - 13 = 2 \sqrt{42}

, which is between

12

and

14

b^{2} - 13 = 1 + 2 \sqrt{33}

, which is between

11

and

13

As the ranges are overlapping, we square these again

(a^{2} - 13)^{2} = 4 \times 42 = 168

(b^{2} - 13)^{2} = 133 + 4 \sqrt{33}

, which is between

153

and

157

Clearly

a > b

Answer:

a

is greater

Note: Squaring and subtraction of variable on the Left Hand Side (LHS) in the expressions above is to make our explanations clear. In the exam, you can write down the operations one below the other as shown below.

(\sqrt{7} + \sqrt{6})^{2}

and

(\sqrt{3} + \sqrt{11})^{2}

⇒

13 + 2 \sqrt{42}

and

14 + 2 \sqrt{33}

⇒

2 \sqrt{42}

and

1 + 2 \sqrt{33}

⇒

168

and

133 + 4 \sqrt{33}

Also, we cannot remove the square roots, add the integers and check. In this case

(7 + 6) < (3 + 11)

. However,

\sqrt{7} + \sqrt{6} > \sqrt{3} + \sqrt{11}

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