Surds

Surds are just expressions with a “𝑛𝑡ℎ root” sign.

Properties of Surds: (For 𝑎, 𝑏 ≥ 0)

 √𝑎𝑏 = √𝑎√𝑏
𝑎 √𝑎
 √ = 𝑛
( √𝑎)𝑛 = 𝑎
𝑏 √𝑏

 (√𝑎)2 = 𝑎 where 𝑎 ≥ 0
 𝑏√𝑎 = √𝑏 2 √𝑎 = √𝑏 2 × 𝑎
Simplification of Surds:

Surds with bigger numbers can be simplified by extracting out a perfect square.

Example:

- √147 = √49 × 3 = √49 × √3 = 7√3

- √32 = √16 × 2 = √16 × √2 = 4√2

Rationalization of Denominator:

When the denominator contains just square roots, we could get rid of the square roots by
multiplying the top and bottom by surds:

1−√3 1−√3 √2 √2(1−√3) √2−√6

= × = =
√2 √2 √2 √2×√2 2
Multiply by 1 (does not change the value of an expression)

When the denominator contains multiple surds/constant and surd terms, we multiply the top
and bottom by the conjugate of the denominator:
Conjugate of “1 − √3” = 1 + √3
2+√3 2+√3 2√3+3√2
= × Multiply by
Conjugate of “√5 − √3” = √5 + √3
2√3−3√2 2√3−3√2 2√3+3√2 Conjugate of
“2√3 − 3√2” Conjugate of “2√5 + 3√3” = 2√5 − 3√3”
2 + √3 2√3 + 3√2
= ×
2√3 − 3√2 2√3 + 3√2

(2 + √3)(2√3 + 3√2) Multiplication of surd conjugates is done using

=
(2√3 + 3√2)(2√3 − 3√2) the algebra formula: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 :

4√3 + 6√2 + 6 + 3√6 (2√3 + 3√2)(2√3 − 3√2) = (2√3)2 − (3√2)2

=
−6 = (4)(3) − (9)(2)
= −6
Rationalization of More Complicated Denominator:

2 2
 =
(3−√2)2 32 −(2)(3)(√2)+(√2)2 Expand out the denominator
2 11 + 6√2
= × Multiply the top & bottom by the conjugate
11 − 6√2 11 + 6√2
2 11 + 6√2 We can multiply the top and bottom by
= ×
11 − 6√2 11 + 6√2 (3 + √2)2

2(11+6√2) 22+12√2 22+12√2

= 2 = =
112 −(6√2) 121−(62 )(√2)2 49

Alternatively:

2 2 2
2 2 (3+√2) 2(3+√2) 2[32 +2(3)(√2)+(√2) ]
= × 2 = 2 = 2
(3−√2)2 (3−√2)2 (3+√2) (3−√2)2 (3+√2) [(3−√2)(3+√2)]

Multiply by the conjugate of 2 2

2
(3 − √2)2 (3 + √2) = [(3 − √2)(3 + √2)]
(3 − √2)2 : Which is “(3 + √2) ”
Using the law of indices
2(11+6√2) 22+12√2
= 2 2
=
49
[32 −(√2) ]

√3
 Conjugate of “(1 + √2) − √3” is: (1 + √2) + √3
1+√2−√3

Multiply top & bottom by the conjugate of the denominator gives:

√3 √3 1+√2+√3 √3(1+√2+√3)
= × =
1+√2−√3 1+√2−√3 1+√2+√3 (1+√2−√3)(1+√2+√3)

Use (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2
2
√3 + √2√3 + (√3) √3 + √6 + 3
= 2 2 =
(1 + √2) − (√3) (1 + 2√2 + 2) − 3

√3 + √6 + 3 √3 + √6 + 3 √2 Multiply by
= = × √2 (NO need
2√2 2√2 √2
Rationalize the denominator to multiply by
√2(√3 + √6 + 3) √6 + √12 + 3√2 2√2)
= =
2(2) 4
√6 + 2√3 + 3√2
=
4
 2 3√54 15 14
Example 1: Find the value of 𝑘 when: ( − − ) = k√3
√2 4 √150 √294

Solution:

2 3√54 15 14
( − − ) = k√3
√2 4 √150 √294 Simplify the surds √54, √150, √294

2 3√9×6 15 14
( − − ) = k√3
√2 4 √25×6 √49×6

2 9√6 15 14
( − − ) = k√3
√2 4 5√6 7 √6 Rationalize the denominators

2 9√6 15√6 14√6

( − − ) = k√3
√2 4 5(6) 7(6)

2 9√6 3√6 2 √6
( − − ) = k√3
√2 4 6 6
Simplify the like terms within the
2 17√6 brackets
( ) = k√3
√2 12

2 17√3√2
( ) = k√3
√2 6

17√3 17
= 𝑘√3 ⟹ 𝑘 =
3 3

Example 2:

2 2 2 2
Express ( ) − + (1 − √3) in the form of 𝑎 + 𝑏√3.
√3+2 3−√3

Solution:

2 2 2 2 Simplify the Surd form

( ) − + (1 − √3)
√3+2 3−√3

4 2 3+√3 2
= 2 − × + (1 − √3)
(√3+2) 3−√3 3+√3

2
4(√3−2) 2(3+√3)
= 2 2 − + (1 − 2√3 + 3)
(√3+2) (√3−2) 9−3
 4(3−2√3+4) 3+√3
= − + (1 − 2√3 + 3)
[(√3+2)(√3−2)]2 3

4(7−2√3) √3 √3
= − (1 + ) + 4 − 2√3 = 28 − 8√3 − 1 − + 4 − 2√3
(3−4)2 3 3

√3
= 31 − 10√3 −
3

31√3 31
= 31 − ⟹ 𝑎 = 31, 𝑏 = −
3 3

Example 3:

2+√3 2 2
2+√3 2−√3
By first expressing in the form of 𝑎 + √𝑏. Find the exact value of ( ) + (2+ 3)
2−√3 2−√3 √

Solution:
2
2+√3 2+√3 2+√3 (2+√3) 7+4√3
= × = = = 7 + 4√3
2−√3 2−√3 2+√3 4−3 1

2 2 2
2+√3 2−√3 1
( ) +( ) = (7 + 4√3) + 2
2−√3 2+√3 (7+4√3)
1
= (49 + 56√3 + 48) +
49 + 56√3 + 48
1
= (97 + 56√3) +
97 + 56√3
1 97 + 56√3
= (97 + 56√3) + ×
97 + 56√3 97 + 56√3
97 + 56√3
= (97 + 56√3) + 2
97 − 562 (3)
73 − 40√3
= (73 + 40√3) +
1
= (73 + 40√3) + (73 − 40√3)
= 146
 If any one of 𝑎, 𝑏, 𝑑 and 𝑒 are NOT rational numbers,
then the “equality of surds” cannot be applied!
Equality of Surds:

If rational numbers 𝑎, 𝑏, 𝑑, 𝑒 are such that:

𝑎 + 𝑏√𝑐 = 𝑑 + 𝑒√𝑐, then: 𝑎 = 𝑑 and 𝑏 = 𝑒

This rule is used in finding unknown constants by comparison of non-surd and surd form
respectively on both sides.

Example 1:

Given that (2 + √3)(3 − 𝑏√3) = 𝑎 + 7√3, find the value of 𝑎 and 𝑏.

Solution: Collate all the non-surds and

surd terms
𝐿. 𝐻. 𝑆 = (2 + √3)(3 − 𝑏√3) = 6 − 2𝑏√3 + 3√3 − 𝑏(√3)(√3)

= (6 − 3𝑏) + √3(3 − 2𝑏)

𝑎 7√3
By comparison (equality of surds) with 𝑅. 𝐻. 𝑆:

6 − 3𝑏 = 𝑎 --- (1)
3 − 2𝑏 = 7 --- (2)

Solving for 𝑏 (from equation (2)) gives: 𝑏 = −2

Substitute 𝑏 = −2 into equation (1) ⟹ 𝑎 = 6 − 3(−2) = 12

Example 2:

2+√10
Find the value of 𝑝 and 𝑞 for which: = 𝑝√2 + 𝑞√5
2√5+3√2
When there is no unknowns for the
Solution: fraction term, FASTEST way is to
rationalize the denominator
We will rationalize the denominator on the 𝐿. 𝐻. 𝑆:

2+√10 2+√10 2√5−3√2 (2+√10)(2√5−3√2)

𝐿. 𝐻. 𝑆 = = × = 2 2
2√5+3√2 2√5+3√2 2√5−3√2 (2√5) −(3√2)
4√5 − 6√20 + 2√50 − 3√20 4√5 − 6(√4)(√5) + 2(√25)(√2) − 3(√4)(√5)
= =
20 − 18 2

4√5−12√5+10√2−6√5 8√5+10√2
Simplify the surds (√20 and √50) = = = 4√5 + 5√2
2 2
𝑞√5 + 𝑝√2
Comparing with 𝑅. 𝐻. 𝑆 ⟹ 𝑞 = 4 and 𝑝 = 5

Example 3:

6+4√3
If = 1 − √3, where 𝑎 and 𝑏 are rational numbers, find the value of 𝑎 and 𝑏.
𝑎+𝑏√3
When there is unknown at the denominator, multiple the
denominator to the other side.

Solution:

6+4√3
= 1 − √3
𝑎+𝑏√3 Multiply the denominator over

6 + 4√3 = (1 − √3)(𝑎 + 𝑏√3)

Expand out the 𝑅. 𝐻. 𝑆:

𝑅. 𝐻. 𝑆 = (1 − √3)(𝑎 + 𝑏√3) = a + b√3 − √3𝑎 − 𝑏(√3)(√3)

= (𝑎 − 3𝑏) + √3(𝑏 − 𝑎)

Thus, by comparison: 6 4√3

𝑎 − 3𝑏 = 6 --- (1) Solving simultaneously gives: 𝑎 = −9, 𝑏 = −5

𝑏 − 𝑎 = 4 ---(2)

Example 4:
Square both sides to get rid of the square root containing the unknown
9
If √𝑝 + 𝑞√7 = , where 𝑝 and 𝑞 are rational numbers, find the values of 𝑝 and 𝑞.
4−√7

Solution:

9 9 2
√𝑝 + 𝑞 √7 = ⟹ 𝑝 + 𝑞 √7 = ( )
4−√7 4−√7 Rationalize the denominator since the
denominator has no unknown
81 (4+√7)2
𝑝 + 𝑞 √7 = ×
(4−√7)2 (4+√7)2

81 (4+√7)2
𝑝 + 𝑞 √7 = ×
(4−√7)2 (4+√7)2
2
81(4+√7) 81(16+8√7+7)
𝑝 + 𝑞 √7 = = = 23 + 8√7
[16−7]2 81
By comparison, 𝑝 = 23 and 𝑞 = 8
Finding unknown in an Equation involving surds:

Squaring an equation in the process of solving for an unknown sometimes introduces

“unwanted” roots (a.k.a “extraneous roots”) ⟹ must check the solution after solving.

Knowing when to square both sides in a “surd equation” is Consider the “simple” equation:
important
(Squaring at the wrong time makes the question 𝑥 − 1 = 2 (𝑥 = 3)
unnecessarily complicated)
If we square both sides of the
equation:
Example 1:
(𝑥 − 1)2 = 4
Solve the equation: √3𝑥 + 4 − 4 = 1
Now we take the square roots on
Isolate the numbers and the surd both sides again:
term and the constants when there is
𝑥 − 1 = ±2 ⟹ 𝑥 = 3 or 𝑥 = −1
only 1 surd

Solution: 𝑥 = −1 is an “extraneous” root

√3𝑥 + 4 − 4 = 1
Isolate the surd on the 𝐿. 𝐻. 𝑆
√3𝑥 + 4 = 5
Square both sides to get rid of the square root
3𝑥 + 4 = 25

3𝑥 = 21 ⟹ 𝑥 = 7 (No extraneous roots)

Example 2:
𝑥
Solve the equation: 2√𝑥 − 1 − 𝑥 + 1 =
2

Solution:
𝑥
2 √𝑥 − 1 − 𝑥 + 1 =
2
Isolate the surd on the 𝐿. 𝐻. 𝑆
3𝑥
2 √𝑥 − 1 = −1
2

2 3𝑥 2 9𝑥 2
[2√𝑥 − 1] = ( 2 − 1) ⟹ 4(𝑥 − 1) = 4
− 3𝑥 + 1

16(𝑥 − 1) = 9𝑥 2 − 12𝑥 + 4 ⟹ 9𝑥 2 − 28𝑥 + 20 = 0

(9𝑥 − 10)(𝑥 − 2) = 0
 9
∴𝑥= or 𝑥 = 2
10
9
𝑥= is not a solution since √𝑥 − 1 will not be defined (we cant square root a negative
10
number)

Substitute 𝑥 = 2 into the original equation:

𝐿. 𝐻. 𝑆 = 2√2 − 1 − 2 + 1 = 1
2
𝑥 = 2 is a solution
𝑅. 𝐻. 𝑆 = 2 = 1

Example 3:
Bring one of the surd to the side of
Solve the equation: 1 = √𝑥 + 2 − √3 − 𝑥. the constant before squaring

Solution:

1 = √𝑥 + 2 − √3 − 𝑥
Bring the Surd form to the other side
√𝑥 + 2 = 1 + √3 − 𝑥
Square both sides
(√𝑥 + 2)2 = [1 + √3 − 𝑥 ] 2

𝑥 + 2 = 1 + 2√3 − 𝑥 + 3 − 𝑥

2𝑥 − 2 = 2√3 − 𝑥
Square both sides again to get rid of square roots
2
(2𝑥 − 2) = 4(3 − 𝑥)

4𝑥 2 − 8𝑥 + 4 = 12 − 4𝑥

4𝑥 2 − 4𝑥 − 8 = 0 ⟹ 𝑥2 − 𝑥 − 2 = 0

(𝑥 − 2)(𝑥 + 1) = 0

𝑥 = 2 or 𝑥 = −1
Check your solution since there are multiple solutions:

Substitute 𝑥 = 2 into the original equation:

𝑅. 𝐻. 𝑆 = √2 + 2 − √3 − 2 = 2 − 1 = 𝐿. 𝐻. 𝑆 ⟹ 𝑥 = 2 is a solution

Substitute 𝑥 = −1 into the original equation:

𝑅. 𝐻. 𝑆 = √2 + (−1) − √3 − (−1) = 1 − 2 = −1 ≠ 𝐿. 𝐻. 𝑆 ⟹ 𝑥 = −1 is NOT a
solution

Example 4:

Without using a calculator, find the values of the integers 𝑎 and 𝑏 for which the solution of
𝑎+√𝑏 𝑎+√𝑏
the equation 𝑥√24 = 𝑥√2 + √6 is . [Why do we not substitute 𝑥 = 11 into the
11
equation? We will end up with an equation
Solution: which we cannot apply “equality of surds”]
𝑎+√𝑏
Since 𝑥 = 11 is a solution to the equation, we shall make 𝑥 the subject first and then
compare with the solution:
Rationalize the denominator
√6
𝑥√24 = 𝑥√2 + √6 ⟹ 𝑥 =
√24−√2

√6 √24+√2 √6(√24+√2)
𝑥= × =
√24−√2 √24+√2 24−2

√6(2√6+√2) 2(6)+√6√2 12+√3√2√2 12+2√3 6+√3

𝑥= = = = =
22 22 22 22 11

6+√3 𝑎+√𝑏
Compare with gives: 𝑎 = 6, 𝑏 = 3
11 11

Problems involving surds:

Example 1:

A cuboid has a square base of side (√3 + 1) m and the height and volume of the cuboid
respectively are 𝑥 𝑚 and (𝑥√48 + √12) 𝑚3 respectively . Find 𝑥, leaving your answer in the
form of 𝑎 + 𝑏√3, where 𝑎 and 𝑏 are integers.

Solution:

Volume of cuboid = (Area of square base)(Height of Cuboid)

𝑥√48 + √12 = 𝑥(√3 + 1)2 Find 𝑥

𝑥√48 + √12 = 𝑥(3 + 2√3 + 1)

𝑥(√48 − 2√3 − 4) = −√12

√12 √12 √12

𝑥=− = − =−
√48−2√3−4 4√3−2√3−4 2√3−4

√12 2√3+4 √12(2√3+4) 2√3(2√3+4) 12+8√3

𝑥=− × =− =− (4)(3)−16
= −
2√3−4 2√3+4 (2√3−4)(2√3+4) −4

∴ 𝑥 = 3 + 2√3
Example 2: (More Challenging)

If 𝑛 is a positive integer, show that:

1
= √𝑛 + 1 − √𝑛
√𝑛 + √𝑛 + 1
1 1 1 1
Hence, find the value of + + +⋯+ .
√25+√26 √26+√27 √27+√28 √99+√100

Solution:

1 √𝑛−√𝑛+1
𝐿. 𝐻. 𝑆 = = = −(√𝑛 − √𝑛 + 1) = √𝑛 + 1 − √𝑛 = 𝑅. 𝐻. 𝑆
√𝑛+√𝑛+1 𝑛−(𝑛+1)

1 1 1 1
+ + + ⋯+
√25+√26 √26+√27 √27+√28 √99+√100
𝑛 = 25 𝑛 = 26 𝑛 = 27 𝑛 = 99

= (√26 − √25) + (√27 − √26) + (√28 − √27) + ⋯ + (√100 − √99)

= √26 − √25
+√27 − √26
+√28 − √27 = √100 − √25
..
.
+√100 − √99

∴ Solution = 10 − 5 = 5