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Surds & Indices

Surds And Indices

MODULES

Basics of Surds
Comparison of Surds
Root of Surds
Indices Rules
Comparing Indices
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Surds & Indices 1
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PRACTICE

Surds & Indices : Level 1
Surds & Indices : Level 2
Surds & Indices : Level 3
ALL MODULES

CAT 2025 Lesson : Surds & Indices - Comparison of Surds

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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+8a = 2 + \sqrt{8}a=2+8​, b=4+3b = 4 + \sqrt{3}b=4+3​ and c=5−2c = 5 - \sqrt{2}c=5−2​. Arrange aaa, bbb and ccc in descending order.

Solution

We can compare two surds at a time.
a−2=8a - 2 = \sqrt{8}a−2=8​ and b−2=2+3b - 2 = 2 + \sqrt{3}b−2=2+3​

2<8<32 < \sqrt{8} < 32<8​<3 and 1<3<21 < \sqrt{3} < 21<3​<2

∴(a−2)\therefore (a - 2)∴(a−2) is between 222 and 333. (b−2)(b - 2)(b−2) is between 333 and 444
∴b>a\therefore b > a∴b>a

b−4=3b - 4 = \sqrt{3}b−4=3​, which is between 111 and 222
c−4=1−2c - 4 = 1 - \sqrt{2}c−4=1−2​, which is between −1-1−1 and 000
∴b>c\therefore b > c∴b>c

a−2=8a - 2 = \sqrt{8}a−2=8​, which is between 222 and 333
c−2=3−2c - 2 = 3 - \sqrt{2}c−2=3−2​, which is between 111 and 222
∴a>c\therefore a > c∴a>c

Answer:
b>a>cb > a > cb>a>c

Alternatively

8\sqrt{8}8​, which is between 222 and 333, can be written as 222.##, where ## are some digits after the decimal point.

Applying this to
aaa, bbb and ccc, we get

a=2+8=2+2a = 2 + \sqrt{8} = 2 + 2a=2+8​=2+2.## = 444.##

b=4+3=4+1b = 4 + \sqrt{3} = 4 + 1b=4+3​=4+1.## = 555.##

c=5−2=5−1c = 5 - \sqrt{2} = 5 - 1c=5−2​=5−1.## = 333.##

∴b>a>c\therefore b > a > c∴b>a>c

Example 3

a=7+6a = \sqrt{7} + \sqrt{6}a=7​+6​ and b=3+11b = \sqrt{3} + \sqrt{11}b=3​+11​. Which of aaa and bbb is greater?

Solution

a2=(7+6)2=13+242a^{2} = (\sqrt{7} + \sqrt{6})^{2} = 13 + 2 \sqrt{42}a2=(7​+6​)2=13+242​
b2=(3+11)2=14+233b^{2} = (\sqrt{3} + \sqrt{11})^{2} = 14 + 2 \sqrt{33}b2=(3​+11​)2=14+233​

a2−13=242a^{2} - 13 = 2 \sqrt{42}a2−13=242​, which is between 121212 and 141414
b2−13=1+233b^{2} - 13 = 1 + 2 \sqrt{33}b2−13=1+233​, which is between 111111 and 131313

As the ranges are overlapping, we square these again
(a2−13)2=4×42=168(a^{2} - 13)^{2} = 4 \times 42 = 168(a2−13)2=4×42=168
(b2−13)2=133+433(b^{2} - 13)^{2} = 133 + 4 \sqrt{33}(b2−13)2=133+433​, which is between 153153153 and 157157157

Clearly
a>ba > ba>b

Answer:
aaa 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.

(7+6)2(\sqrt{7} + \sqrt{6})^{2}(7​+6​)2 and (3+11)2(\sqrt{3} + \sqrt{11})^{2}(3​+11​)2
⇒
13+24213 + 2 \sqrt{42}13+242​ and 14+23314 + 2 \sqrt{33}14+233​
⇒
2422 \sqrt{42}242​ and 1+2331 + 2 \sqrt{33}1+233​
⇒
168168168 and 133+433133 + 4 \sqrt{33}133+433​

Also, we cannot remove the square roots, add the integers and check. In this case
(7+6)<(3+11)(7 + 6) < (3 + 11)(7+6)<(3+11). However, 7+6>3+11\sqrt{7} + \sqrt{6} > \sqrt{3} + \sqrt{11}7​+6​>3​+11​

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