Real Numbers Class 9th Mathematics Part I MHB Solution
Practice Set 2.3
Question 1.
State the order of the surds given below.
i. 
ii. 
iii. 
iv. 
v. 
Answer:
In 
, n is called the order of the surd.
Therefore,
i.
In this, the order of surd is 3.
ii.
In this, the order of surd is 5.
In this, the order of surd is 4.
iv.
In this, the order of surd is 2.
v.
In this, the order of surd is 3
Question 2.
State which of the following are surds. Justify.
i. 
ii. 
iii. 
iv. 
v. 
vi. 
Answer:
Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers.
Therefore,
i.
It is a surd ∵ it cannot be expressed as a rational number.
ii.
It is a surd ∵ it cannot be expressed as a rational number.
iii.
It is a surd ∵ it cannot be expressed as a rational number.
iv.√256 = √162 = 16
It is not a surd ∵ it is a rational number.
v.
It is not a surd ∵ it is a rational number.
vi. 
It is a surd ∵ it cannot be expressed as a rational number.
Question 3.
Classify the given pair of surds into like surds and unlike surds.
i. √52, 5√13
ii. √68, 5√3
iii. 4√18, 7√2
iv. 19√12, 6√3
v. 5√22, 7√33
vi. 5√5, √75
Answer:
Two or more surds are said to be similar or like surds if they have the same surd-factor.
And,
Two or more surds are said to be dissimilar or unlike when they are not similar.
Therefore,
i. √52, 5√13
√52 = √(2×2×13) = 2√13
5√13
∵ both surds have same surd-factor i.e., √13.
∴ they are like surds.
ii. √68, 5√3
√68 = √(2×2×17) = 2√17
5√3
∵ both surds have different surd-factors √17 and √3.
∴ they are unlike surds.
iii. 4√18, 7√2
4√18 = 4√(2×3×3) = 4×3√2 = 12√2
7√2
∵ both surds have same surd-factor i.e., √2.
∴ they are like surds.
iv. 19√12, 6√3
19√12 = 19√(2×2×3) = 19×2√3 = 38√3
6√3
∵ both surds have same surd-factor i.e., √3.
∴ they are like surds.
v. 5√22, 7√33
∵ both surds have different surd-factors √22 and √33.
∴ they are unlike surds.
vi. 5√5, √75
5√5
√75 = √(5×5×3) = 5√3
∵ both surds have different surd-factors √5 and √3.
∴ they are unlike surds.
Question 4.
Simplify the following surds.
i. √27
ii. √50
iii. √250
iv. √112
v. √168
Answer:
i. 
ii. 
iii. 
iv. 
v. 
Question 5.
Compare the following pair of surds.
i. 7√2, 5√3
ii. √247, √274
iii. 2√7, √28
iv. 5√5, 7√2
v. 4√42, 9√2
vi. 5√3, 9
vii. 7, 2√5
Answer:
i. 7√2 , 5√3
(7√2)2 = 7 × 7 × √2 × √2
⇒ (7√2)2 = 49 × 2
⇒ (7√2)2 = 98
And
(5√3)2 = 5 × 5 × √3 × √3
⇒ (5√3)2 = 25 × 3
⇒ (5√3)2 = 75
Clearly,
98 > 75
∴ 7√2 > 5√3
ii. √247, √274
(√247)2 = 247
And
(√274)2 = 274
Clearly,
247 < 274
∴ √247 < √274
iii. 2√7, √28
(2√7)2 = 2 × 2 × √7 × √7
⇒ (2√7)2 = 4 × 7
⇒ (2√7)2 = 28
And
(√28)2 = 28
Clearly,
28 = 28
∴ 2√7 = √28
iv. 5√5, 7√2
(5√5)2 = 5 × 5 × √5 × √5
⇒ (5√5)2 = 25 × 5
⇒ (5√5)2 = 125
And
(7√2)2 = 7 × 7 × √2 × √2
⇒ (7√2)2 = 49 × 2
⇒ (7√2)2 = 98
Clearly,
125 = 98
∴ 5√5= 7√2
v. 4√42, 9√2
(4√42)2 = 4 × 4 × √42 × √42
⇒ (4√42)2 = 16 × 42
⇒ (4√42)2 = 672
And
(9√2)2 = 9 × 9 × √2 × √2
⇒ (9√2)2 = 81 × 2
⇒ (9√2)2 = 162
Clearly,
672 > 162
∴ 4√42 > 9√2
vi. 5√3, 9
(5√3)2 = 5 × 5 × √3 × √3
⇒ (5√3)2 = 25 × 3
⇒ (5√3)2 = 75
And
(9)2 = 9 × 9
⇒ (9)2 = 81
Clearly,
75 < 81
∴ 5√3 < 9
vii. 7, 2√5
(2√5)2 = 2 × 2 × √5 × √5
⇒ (2√5)2 = 4 × 5
⇒ (2√5)2 = 20
And
(7)2 = 7 × 7
⇒ (7)2 = 49
Clearly,
49 > 20
∴ 7 > 2√5
Question 6.
Simplify.
i. 5√3 + 8√3
ii. 9√5 – 4√5 + √125
iii. 7√48 – √27 – √3
iv. 
Answer:
i. 5√3 + 8√3
5√3 + 8√3 = (5 + 8)√3
⇒ 5√3 + 8√3 = 13√3
ii. 9√5 – 4√5 + √125
9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)
⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)
⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + 5√5
⇒ 9√5 – 4√5 + √125 = (9 – 4 + 5)√5
⇒ 9√5 – 4√5 + √125 = 10√5
iii. 7√48 – √27 – √3
7√48 – √27 – √3 = 7√(2 × 2 × 2 × 2 × 3) – √(3 × 3 × 3) – √3
⇒ 7√48 – √27 – √3 = 7×4√3 – 3√3 – √3
⇒ 7√48 – √27 – √3 = 28√3 – 3√3 – √3
⇒ 7√48 – √27 – √3 = (28 – 3 – 1)√3
⇒ 7√48 – √27 – √3 = 24√3
iv. 
Question 7.
Multiply and write the answer in the simplest form.
i. 3√12 × √18
ii. 3√12 × 7√15
iii. 3√8 × √5
iv. 5√8 × 2√8
Answer:
i. 3√12 × √18
3√12 × √18 = 3√(2 × 2 × 3) × √(2 × 3 × 3)
⇒3√12 × √18 = 3 × 2√3 × 3√2
⇒ 3√12 × √18 = 6√3 × 3√2
⇒ 3√12 × √18 = 18√6
ii. 3√12 × 7√15
3√12 × 7√15 = 3√(2 × 2 × 3) × 7√(3 × 5)
⇒3√12 × 7√15 = 3 × 2√3 × 7√(3 × 5)
⇒3√12 × 7√15 = 3 × 2 × 7 × √(3 × 3 × 5)
⇒3√12 × 7√15 = 3 × 2 × 7 × 3√5
⇒ 3√12 × 7√15 = 126√5
iii. 3√8 × √5
3√8 × √5 = 3√(2 × 2 × 2) × √5
⇒3√8 × √5 = 3 × 2√2 × √5
⇒ 3√8 × √5 = 3 × 2 ×√(2 × 5)
⇒ 3√8 × √5 = 6√10
iv. 5√8 × 2√8
5√8 × 2√8 = 5√(2 × 2 × 2) × 2√(2 × 2 × 2)
⇒5√8 × 2√8 = 5 × 2√2 × 2 × 2√2
⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 ×√(2 × 2)
⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 × 2
⇒ 5√8 × 2√8 = 80
Question 8.
Divide, and write the answer in simplest form.
i. √98 ÷ √2
ii. √125 ÷ √50
iii. √54 ÷ √27
iv. √310 ÷ √5
Answer:
i. √98 ÷ √2
⇒√98 ÷ √2 = 7
ii. √125 ÷ √50
iii. √54 ÷ √27
⇒√54 ÷ √27 = √2
iv. √310 ÷ √5
⇒ √310 ÷ √5 = √62
Question 9.
Rationalize the denominator.
i. 
ii. 
iii. 
iv. 
v. 
Answer:
i. We know that √5 × √5 = 5, ∴ to rationalize the denominator of 
multiply both numerator and denominator by √5.
ii. We know that √14 × √14 = 14, ∴ to rationalize the denominator of 
multiply both numerator and denominator by √14.
iii. We know that √7 × √7 = 7, ∴ to rationalize the denominator of 
multiply both numerator and denominator by √7.
iv. We know that √3 × √3 = 3, ∴ to rationalize the denominator of 
multiply both numerator and denominator by √3.
v. We know that √3 × √3 = 3, ∴ to rationalize the denominator of 
multiply both numerator and denominator by √3.
