RATIONAL AND
IRRATIONAL NUMBERS
Q.1. Without actual division find which of the following rationals are
terminating decimal:
9
7
121
37
(i)
(ii)
(iii)
(iv)
25
12
125
78
9
Ans. (i) In
, the prime factors of denominator 25 are 5, 5. Thus it is terminating decimal.
25
7
, the prime factors of denominator 12 are 2, 2 and 3. Thus it is not
(ii) In
12
terminating decimal.
121
(iii) In
, the prime factors of denominator 125 are 5, 5 and 5. Thus it is
125
terminating decimal.
37
(iv) In
, the prime factors of denominator 78 are 2, 3 and 13. Thus it is not
78
terminating decimal.
Q.2. Represent each of the following as a decimal number.
4
5
31
(i)
(ii) 2
(iii) 5
15
12
55
4
, using long division method:
Ans. (i) In
15
0.266...
15 4.0000
30
100
90
100
90
10
4
Hence,
= 0.266... = 0.26.
15
Math Class IX
1
Question Bank
31
5
, using long division method: (iv) In 5 , using long division method:
55
12
0.5636363
0.4166...
55 31.0000000
12 5.000
275
48
350
20
330
12
200
80
165
72
350
80
330
72
200
8
165
5
Hence, 2 = 2.4166... = 2.416.
350
12
330
200
165
35
31
Hence, 5 = 5.5636363... = 5.563.
55
p
Q.3. Express each of the following as a rational number in the form of
,
q
where q ≠ 0 .
(iii) In 2
(i) 0.6
(ii) 0.43
(iii) 0.227
Ans. (i) Let x = 0.6 = 0.6666
...(i)
Multiplying both sides of eqn. (i) by 10, we get
...(ii)
10 x = 6.6666
Subtracting eqn. (i) from eqn. (ii), we get
10 x = 6.6666
x = 0.6666
(iv) 0.2104
6 2
2
= Hence, required fraction = .
9 3
3
(ii) Let x = 0.43 = 0.43434343
Multiplying both sides of eqn. (i) by 100, we get
100 x = 43.434343
Subtracting eqn. (i) from eqn. (ii), we get
9x = 6
Math Class IX
⇒x=
2
...(i)
...(ii)
Question Bank
100 x = 43.434343
x = 0.434343
99 x = 43
⇒ 99 x = 43 ⇒ x =
Hence, required fraction
43
99
p 43
= .
q 99
...(i)
(iii) Let x = 0.227 = 0.2272727...
Multiplying both sides of eqn. (i) by 10, we get
10 x = 2.272727...
Multiplying both sides of eqn. (ii) by 100, we get
1000 x = 227.272727...
Subtracting eqn. (ii) from (iii), we get
1000 x = 227.272727...
10 x = 2.272727...
990 x = 225
⇒ 990 x = 225 ⇒ x =
Hence, required fraction
...(ii)
...(iii)
225 5
=
990 22
p 5
= .
q 22
(iv) Let x = 0.2104 = 0.2104104104...
Multiplying both sides of eqn. (i) by 10, we get
10 x = 2.104104104...
Multiplying both sides of eqn. (ii) by 1000, we get
10000 x = 2104.104104104...
Subtracting eqn. (ii) from (iii), we get
10000 x = 2104.104104104
10 x =
2.104104104
⇒
9990 x = 2102
Hence, required fraction =
Math Class IX
...(i)
...(ii)
...(iii)
9990 x = 2102 ⇒ x =
2102 1051
=
9990 4995
1051
.
4995
3
Question Bank
Q.4. Express each of the following as a vulgar fraction.
(i) 3.146
(ii) 4.324
Ans. Let x = 3.146 = 3.146146
Multiplying both sides of eqn. (i) by 1000, we get
1000 x = 3146.146146146
Subtracting eqn. (i) from eqn. (ii), we get
1000 x = 3146.146146146
x = 3.146146146
999 x = 3143
...(i)
...(ii)
⇒ 999 x = 3143 ⇒ x =
Hence, required vulgar fraction =
3143
999
3143
.
999
(ii) Let x = 4.324 = 4.324242424
Multiplying both sides of eqn. (i) by 10, we get
10 x = 43.24242424
Multiplying both sides of eqn. (ii) by 100, we get
1000 x = 4324.24242424
Subtracting eqn. (ii) from eqn. (iii), we get
1000 x = 4324.24242424
10 x = 43.24242424
990 x = 4281
⇒ 990 x = 4281 ⇒ x =
...(i)
...(ii)
...(iii)
4281 1427
=
990
330
1427
.
330
Q.5. Insert one rational number between:
3
7
(i) and
(ii) 8 and 8.04
5
9
Ans. If a and b are two rational numbers, then between these two numbers, one
(a + b )
.
rational number will be
2
3
7
Required rational number between and
5
9
3 31 7
1  3 7  1  27 + 35  1 62 31
=  + = 
 = 2 × 45 = 45 ∴ 5 < 45 < 9 .
2  5 9  2  45 
Hence, required vulgar fraction =
Math Class IX
4
Question Bank
(ii) Required rational number between 8 and 8.04
1
1
= (8 + 8.04) = (16.04) = 8.02 ∴ 8 < 8.02 < 8.04
2
2
3
1
Q.6. Insert two rational numbers between and 1
4
5
3
1
3
6
3 13 6 6
3 1  15 + 24  6
Ans.
and 1 ⇒ and ⇒ <  +  < ⇒ < 
<
4
5
4
5
4 24 5 5
5 2  20  5
3 39 6
3 1  39  6
<
⇒ <  < ⇒ <
4 40 5
4 2  20  5
3 39 1  39 6  6
3 39 1  39 + 48  6
⇒ <
<  + < ⇒ <
< 
<
4 40 2  40 5  5
4 40 2  40  5
3 39 87 6
3 39 1  87  6
<
<
⇒ <
<  < ⇒ <
4 40 80 5
4 40 2  40  5
39
87
Hence, required rational numbers are
and
.
40
80
Q.7. Insert three rational numbers between
1
3
(i) 4 and 5
(ii) and
2
5
1
2
(iii) 4 and 4.5
(iv) 2 and 3
3
3
1
1
(v) − and
2
3
Ans. (i) The given numbers are 4 and 5.
As,
4<5
1 4+5
9
⇒4< 
<5⇒ 4< <5
2 1 
2
⇒ 4 < 4.5 < 5
1
9 9
Again, 4 <  4 +  <
2
2 2
⇒ 4 < 4.25 < 4.5
1
Again, 4.5 < 5 ⇒ 4.5 < (4.5 + 5) < 5 ⇒ 4.5 < 4.75 < 5
2
∴ From eqn. (i), (ii) and (iii), we get 4 < 4.25 < 4.5 < 4.75 < 5.
Thus, required rational numbers between 4 and 5 are 4.25, 4.75 and 4.5.
Math Class IX
5
...(i)
...(ii)
...(iii)
Question Bank
1
3
and
2
5
1 3
1 1 1 3 3
1 15+ 6 3
As,
< ⇒ <  + < ⇒ < 
<
2 5
2 2 2 5 5
2 2  10  5
1 11 3
1 1  11  3
<
⇒ <  < ⇒ <
2 20 5
2 2  10  5
1 1  1 11  3
1 1  21  3
Again, <  +  < ⇒ <   <
2 2  2 20  5
2 2  20  5
1 21 3
<
<
...(ii)
2 40 5
11 3
11 1  11 3  3
1 1  23  3
Again,
< ⇒
<  + < ⇒ <  <
20 5
20 2  20 5  5
2 2  20  5
1 23 3
⇒ <
<
...(iii)
2 40 5
From eqn. (i), (ii) and (iii), we get
1 21 11 23 3
<
<
<
<
2 40 20 40 15
1
3
21 11
23
Thus, required rational numbers between and are
and
,
.
2
5
40 20
40
(iii) The given numbers are 4 and 4.5
1
As 4 < 4.5 ⇒ 4 < (4 + 4.5) < 4.5
2
...(i)
⇒ 4 < 4.25 < 4.5
1
⇒ 4 < (4 + 4.25) < 4.25 ⇒ 4 < 4.125 < 4.25
...(ii)
2
Again, 4.25 < 4.5
1
⇒ 4.25 < (4.25 + 4.5) + 4.5 ⇒ 4.25 < 4.375 < 4.5
...(iii)
2
From eqn. (i), (ii) and (iii), we have 4 < 4.125 < 4.25 < 4.375 < 4.5
Thus, required rational numbers between 4 and 4.5 are 4.125, 4.25 and 4.375.
(ii) The given numbers are
Math Class IX
6
Question Bank
1
2
7
11
(iv) The given numbers are 2 and 3 i.e., and .
3
3
3
3
7 11
7 1  7 11  11
As < ⇒ <  +  <
3 3
3 23 3  3
7 18 11
7 1  18  11
⇒ <  < ⇒ < <
3 6
3
3 2 3  3
7
11
⇒ <3<
...(i)
3
3
7 1 7 3
Again, <  +  < 3
3 2 3 1
7 8
< <3
...(ii)
3 3
11
Again, 3 <
3
1  11  11
1  20  11
3 < 3+  < ⇒ 3 <   <
2
3 3
2 3  3
10 11
3< <
...(iii)
3
3
From eqn. (i), (ii) and (iii), we get
7 8
10 11
< <3<
< .
3 3
3
3
1
2
7
11
8
10
Thus required rational numbers between 2 and 3 i.e., and
are , 3 and
.
3
3
3
3
3
3
1
2
Q.8. Find the decimal representation of
and . Deduce from the decimal
7
7
1
representation of , without actual calculation, the decimal representa7
3 4 5
6
tion of , , and .
7 7 7
7
Math Class IX
7
Question Bank
1
using long division method.
7
1
Thus decimal representation of = 0.142857
7
2
1
⇒ Decimal representation of = 2 × = 2 × 0.142857 = 0.285714
7
7
3
1
⇒ Decimal representation of = 3 × = 3 × 0.142857 = 0.428571
7
7
4
1
⇒ Decimal representation of = 4 × = 4 × 0.142857 = 0.571428
7
7
5
1
⇒ Decimal representation of = 5 × = 5 × 0.142857 = 0.714285
7
7
6
1
⇒ Decimal representation of = 6 × = 6 × 0.142847 = 0.857142
7
7
Ans. Decimal representation of
0.142871
7 1.000000
7
30
28
20
14
60
56
40
35
50
49
10
7
3
Q.9. State, whether the following numbers are rational or irrational:
(
(i) 2 + 2
(
Ans. (i) 2 + 2
)
)
2
2
(
)(
(ii) 5 + 5 5 − 5
)
= 4 + 2 + 2× 2× 2 = 6 + 4 2
Hence, it is an irrational number.
(
(iii) 5 + 5
) ( 5 − 5 ) = ( 5 )2 − ( 5 )
2
[Using (a + b)(a − b) = a 2 − b 2 ]
= 25 − 5 = 20
Hence, it is a rational number.
Q.10. Given
universal
3
3 3
4
2
set = {–6, – 5 , – 4, – , – , 0, , 1, 1 , 8, 3.01, π, 8.47}
4
5 8
5
3
From the given set find:
(i) Set of rational numbers
(ii) Set of irrational numbers
(iii) Set of integers
(iv) Set of non-negative integers
Ans. The given universal set is
3
3 3
4
2
{−6, − 5 , − 4, − , − , 0, , 1, 1 , 8, 3.01, π , 8.47}
4
5 8
5
3
Math Class IX
8
Question Bank
(i) Set of rational numbers
3
3 3
4
2
= {−6, − 5 , − 4, − , − , 0, , 1, 1 , 3.01, 8.47}
4
5 8
5
3
(ii) Set of irrational numbers = { 8, π }
(iii) Set of integers = {−6, − 4, 0, 1}
(iv) Set of non-negative integers = {0, 1}
Q.11. Use division method to show that 3 and 5 are irrational numbers.
Ans. 3 = 1.73205...
It is non-terminating and non-recurring decimals.
1.73205
1 3.00 00 00 00 00
∴
3 is an irrational number.
1
27 200
189
343 1100
1029
3462
7100
6924
346405
1760000
1732025
28975
5 = 2.2360679...
It is non terminating and non recurring decimals.
2.2360679
2 5.00 00 00 00 00 00
∴ 5 is an irrational number.
1
42 100
84
443 1600
1329
4466
27100
26796
447206
3040000
2683236
4472127
35676400
31304889
44721349
437151100
402492141
34658959
Math Class IX
9
Question Bank
Q.14. Show that
5 is not a rational number.
p
Ans. Let 5 is a rational number and let 5 = .
q
Where p and q have no common factor and q ≠ 0.
Squaring both sides, we get
( 5)
2
2
 p
p2
=   = 5 = 2 ⇒ p 2 = 5q 2
q
q
...(i)
⇒ p 2 is a multiple of 5 ⇒ p is also multiple of 5
Let p = 5m for some positive integer m.
...(ii)
p 2 = 25m 2
From eqn. (i) and (ii), we get
5q 2 = 25m 2 ⇒ q 2 = 5m 2 ∴ q 2 is multiple of 5 ⇒ q is multiple of 5
Thus, p and q both are multiple of 5.
This shows that 5 is a common factor of p and q . This contradicts the hypothesis
that p and q have no common factor, other than 1.
∴ 5 is not a rational number.
Q.13. Show that:
( 3 + 7 ) is an irrational number
(ii) ( 3 + 5 ) is an irrational number.
Ans. (i) Let ( 3 + 7 ) is a rational number.
2
Then square of given number i.e., ( 3 + 7 ) is rational.
2
⇒ ( 3 + 7 ) is rational
2
2
⇒ ( 3 ) + ( 7 ) + 2 3 × 7 = 3 + 7 + 2 21 = (10 + 2 21 ) is rational
But, (10 + 2 21 ) being the sum of a rational and irrational is irrational. This
contradiction arises by assuming that ( 3 + 7 ) is rational number.
(i)
Hence,
3 + 7 is an irrational number.
Math Class IX
10
Question Bank
(ii) Let
(
)
3 + 5 is a rational number.
Then square of given number i.e.,
(
3+ 5
)
2
is rational.
2
( 3 + 5 ) is rational
2
2
⇒ ( 3 ) + ( 5 ) + 2 3 × 5 = 3 + 5 + 2 15 = ( 8 + 2 15 ) rational.
But, ( 8 + 2 15 ) being the sum of a rational and irrational it is irrational. This
contradiction arises by assuming that ( 3 + 5 ) is rational.
Hence, ( 3 + 5 ) is irrational number.
⇒
Q.14. Use method of contradiction to show that
Ans. (i) Now, Let 3 is a rational number
p
(where q ≠ 0)
3=
q
Then,
( 3)
⇒ 3=
2
p2
q
2
 p
= 
q
3 is an irrational number.
2
⇒ p 2 = 3q 2
∴ p 2 is divisible by 3 as 3q 2 is divisible by 3.
⇒ p is divisible by 3.
Let p = 3r
Then p 2 = 9r 2
2
⇒ 3q = 9r
2
(On squaring both sides)
2
⇒ q = 3r 2
∵ 3r 2 is also divisible by 3.
∴ q is divisible by 3.
From (i) and (ii), we get
p
is divisible by 3.
q
Math Class IX
...(i)
...(ii)
11
Question Bank
∴ p and q have 3 as their common factor but
have no common factor. ∴
p
is a rational number i.e. p and q
q
p
is not rational. So
q
3 is not rational. Hence,
3 is irrational number.
Q.15. Insert three irrational numbers between 0 and 1.
Ans. Three irrational numbers between 0 and 1 can be
0 < 0.1011001110001111... < 0.1010011000111... < 0.202002000200002... < 1
Q.16. Rationalise the denominator and simplify.
1
6
(i)
(ii)
5+ 2
3– 5
(iii)
1
(iv)
2 5– 3
7+3 5
3+ 5
–
7–3 5
3– 5
1
1
(vi)
1+ 5 + 3
6 + 5 – 11
1
Ans. (i)
3− 5
Multiplying numerator and denominator by 3 + 5 , we get
1
3+ 5
=
3− 5
3− 5 3+ 5
(v)
(
=
(ii)
)(
)
3+ 5
( 3) 2 − (
5
)
2
=
3+ 5 3+ 5
=
9−5
4
6
5+ 2
Multiplying numerator and denominator by
6
(
(
5− 2
5+ 2
Math Class IX
)(
)
5−
5 − 2 , we get
) {∵ (a + b)(a − b) = a2 − b2}
2 ) ( 5 )2 − ( 2 )2
6( 5 − 2 ) 6( 5 − 2 )
=
=
= 2( 5 − 2 )
5− 2
3
6
=
(
{∵ (a + b)(a − b ) = a 2 − b 2 }
5− 2
12
Question Bank
(iii)
1
2 5− 3
Multiplying numerator and denominator by 2 5 + 3 , we get
(
)
2 5+ 3
1
2 5+ 3
=
=
2
2 5− 3
2 5− 3 2 5+ 3
2 5 − 3
(
(iv)
)(
) (
) ( )
2
=
2 5+ 3
17
7+3 5 7−3 5
−
3+ 5
3− 5
(7 + 3 5) (3 − 5) − (7 − 3 5) (3 + 5)
=
(3 + 5) (3 − 5)
=
21 − 7 5 + 9 5 − 15 − 21 − 7 5 + 9 5 + 15
(3)2 − ( 5)2
21 + 2 5 − 15 − 21 + 2 5 + 15 4 5
=
= 5
9−5
4
1
(v)
1+ 5 + 3
Multiplying numerator and denominator by 1 − ( 5 + 3) , we get
=
1
1
1 − ( 5 + 3)
=
×
1 + 5 + 3 1 + ( 5 + 3) 1 − ( 5 + 3)
=
1 − ( 5 + 3)
2
(1) − ( 5 + 3)
2
=
1− 5 − 3
1 − (5 + 3 + 2 15)
1− 5 − 3 1− 5 − 3
5 + 3 −1
=
=
1 − 8 − 2 15 −7 − 2 15
7 + 2 15
Multiplying numerator and denominator by 7 − 2 15 , we get
=
5 + 3 −1
5 + 3 − 1 (7 − 2 15)
=
×
7 + 2 15
7 + 2 15 (7 − 2 15)
Math Class IX
=
7 5 + 7 3 − 7 − 2 75 − 2 45 + 2 15
49 − 4 (15)
=
7 5 + 7 3 − 7 − 2 × 5 3 − 2 × 3 5 + 2 15
−11
13
Question Bank
7 5 + 7 3 − 7 − 10 3 − 6 5 + 2 15
−11
5 − 3 3 − 7 + 2 15 −(7 − 5 + 3 3 − 2 15)
=
=
−11
−11
7 − 5 + 3 3 − 2 15
=
11
=
(vi)
1
6 + 5 − 11
Multiplying numerator and denominator by
6 + 5 + 11 , we get
1
[( 6 + 5) + 11]
=
6 + 5 − 11 [( 6 + 5) − 11][ 6 + 5) + 11]
=
6 + 5 + 11
( 6 + 5)2 − ( 11)2
=
6 + 5 + 11
6 + 5 + 2 30 − 11
6 + 5 + 11
2 30
Multiplying numerator and denominator by
=
=
30 , we get
( 6 + 5 + 11) 30
2 30 × 30
180 + 150 + 330
2 × 30
36 × 5 + 25 × 6 + 330
=
60
6 5 + 5 6 + 330
=
60
=
3 –1
= a + b 3, find the value of a and b .
3 +1
3 −1
Ans.
= a+b 3
3 +1
Multiplying both sides numerator and denominator of L.H.S. by ( 3 − 1),
we get
3 − 1 ( 3 − 1) ( 3 − 1)
⇒
=
3 + 1 ( 3 + 1) ( 3 − 1)
Q.17. If
Math Class IX
14
Question Bank
3 + 1 − 2 3 ×1
3 −1
4−2 3
⇒
2
⇒ 2− 3
3 −1
But
= a + b 3, so 2 − 3 = a + b 3
3 +1
Comparing both sides
⇒ a = 2 and b = −1
⇒
3+ 2
= a + b 2 , find the value of a and b.
3- 2
3+ 2
Ans.
= a+b 2
3− 2
Multiplying numerator and denominator of L.H.S. by 3 + 2 , we get
3 + 2 (3 + 2) (3 + 2)
=
3 − 2 (3 − 2) (3 + 2)
Q.18. If
⇒
(3 + 2)2
2
(3) − ( 2)
2
9 + 2 + 2×3 2
11 + 6 2
⇒
9−2
7
⇒
11 6 2
3+ 2
11 6 2
+
But
+
= a +b 2
= a + b 2, so
7
7
7
7
3− 2
11
6
Comparing both sides, a = , b =
7
7
Q.19. Simplify:
⇒
(i)
22
+
17
(ii)
2
3
–
(iii)
19
2 3 +1 2 3 –1
6– 2
6+ 2
3 2–2 3
22
17
Ans. (i)
+
2 3 +1 2 3 −1
By rationalising the denominator of each term, we get
22
17
22
2 3 −1
17
2 3 +1
+
=
×
+
×
2 3 + 1 2 3 − 1 2 3 + 1 2 3 −1 2 3 − 1 2 3 + 1
44 3 − 22
34 3 + 17
=
+
2
2
2 3 − (1) 2
2 3 − (1)2
(
Math Class IX
)
(
15
+
1
3 2+2 3
)
Question Bank
44 3 − 22 34 3 + 17 44 3 − 22 34 3 + 17
+
=
+
12 − 1
12 − 1
11
11
44 3 − 22 + 34 3 + 17 78 3 − 5
=
=
11
11
2
3
(ii)
−
6− 2
6+ 2
By rationalising the denominator of each term, we get
2
3
2
6+ 2
3
6− 2
−
=
×
−
×
6− 2
6+ 2
6− 2
6+ 2
6+ 2
6− 2
2× 6 + 2
3× 6 − 6
=
−
2
2
( 6) − ( 2)
( 6) 2 − ( 2) 2
=
2× 2×3 + 2
3× 3× 2 − 6
−
6−2
6−2
2 3 + 2 − (3 2 − 6) 2 3 + 2 − 3 2 + 6
=
=
4
4
18
1
(iii)
+
3 2 −2 3 5 2 +2 3
By rationalising the denominator of each term, we get
18
1
18
3 2+2 3
1
5 2−2 3
+
=
×
+
×
3 2 −2 3 5 2 +2 3 3 2 −2 3 3 2 +2 3 5 2 +2 3 5 2 −2 3
54 2 + 36 3
5 2 −2 3
=
+
2
2
(3 2) − (2 3)
(5 2) 2 − (2 3) 2
=
54 2 + 36 3 5 2 − 2 3
+
18 − 12
38
19(54 2 + 36 3) + 3(5 2 − 2 3)
=
114
1026 2 + 684 3 + 15 2 − 6 3
=
114
1041 2 + 678 3 1041 2 678 3
=
=
+
114
114
114
347 2 113 3
=
+
38
19
=
Math Class IX
16
Question Bank
5 –2
Q.20. If x =
5 +2
5 +2
5 –2
; find : x 2 + y 2 + xy
(ii) y 2
(i) x 2
Ans. (i) x =
and y =
(iv) x2 + y 2 + xy
(iii) xy
5−2
5−2
5−2
=
×
5+2
5+2
5−2
( 5 − 2) 2 5 + 4 − 4 5
=
=
=9−4 5 ∴ x =9−4 5
5− 4
1
Squaring both sides, we get
⇒ x 2 = (9 − 4 5)2 = 81 + 16(5) − 72 5 = 81 + 80 − 72 5 = 161 − 72 5
5+2
5 + 2 5+ 4+ 4 5
(ii) y =
×
=
=9+4 5
5− 4
5−2
5+2
y =9+4 5
Squaring both sides, we get
∴ y 2 = (9 + 4 5)2 = 81 + 80 + 2 × 9 × 4 5 = 161 + 72 5
(iii) xy = (9 − 4 5) (9 + 4 5) = 81 − 80 = 1
(iv) x 2 + y 2 + xy = 161 − 72 5 + 161 + 72 5 + 1 = 323
Q.21. Write down the values of:
3

(i) 
2
2

2
(i) (5 + 3)2
(iii) ( 6 – 3)2
(iv) ( 5 + 6 )2
2
3
3
9
9
9
3

Ans. (i) 
2 =
2×
2 = ( 2) 2 = × 2 =
2
2
4
4
2
2

2
2
2
(ii) (5 + 3) = (5) + ( 3) + 2(5) ( 3)
[using (a + b)2 = a 2 + b2 + 2ab]
= 25 + 3 + 10 3
= 28 + 10 3
(iii) ( 6 − 3) 2 = ( 6) 2 + (3) 2 − 2 × 6 × 3
[using ( a − b ) 2 = a 2 + b 2 − 2 ab]
= 6 + 9 − 6 6 = 15 − 6 6
(iv) ( 5 + 6) = ( 5)2 + ( 6)2 + 2 × 5 × 6
[using (a + b)2 = a 2 + b2 + 2ab]
= 5 + 6 + 2 30 = 11 + 2 30
Math Class IX
17
Question Bank
Q.22. Rationalize the denominator of:
3– 2
3+ 2
(i)
(ii)
5+ 3
(iii)
(iv)
7+ 5
7– 5
2 5+3 2
5– 3
2 5-3 2
3− 2
Ans. (i)
3+ 2
Multiplying numerator and denominator by 3 − 2 , we get
3− 2
3− 2
( 3 − 2)2
×
=
3+ 2
3 − 2 ( 3)2 − ( 2)2
=
( 3)2 + ( 2)2 − 2 × 3 × 2
3−2
3+ 2−2 6
=
=5−2 6
1
7+ 5
(ii)
7− 5
Multiplying numerator and denominator by
=
7 + 5 , we get
2
7+ 5
7+ 5
( 7 + 5)
×
=
7− 5
7 + 5 ( 7 )2 − ( 5)2
=
( 7)2 + ( 5)2 + 2 × 7 × 5
=
7−5
7 + 5 + 2 35 12 + 2 35 2 (6 + 35)
=
=
=
2
2
2
= 6 + 35
5+ 3
(iii)
5− 3
Multiplying numerator and denominator by
5+ 3
5+ 3
( 5 + 3)2
×
=
5− 3
5 + 3 ( 5)2 − ( 3) 2
=
=
5 + 3, we get
( 5)2 + ( 3)2 + 2 × 3 × 5
5−3
Math Class IX
18
Question Bank
5 + 3 + 2 15 8 + 2 15 2 (4 + 15)
=
=
= 4 + 15
2
2
2
2 5+3 2
(iv)
2 5 −3 2
Multiplying numerator and denominator by 2 5 + 3 2 , we get
=
2 5+3 2 2 5 +3 2 2 5 +3 2
(2 5 + 3 2)2
=
×
=
2 5 + 3 2 2 5 − 3 2 2 5 + 3 2 (2 5)2 − (3 2)2
(2 5)2 + (3 2)2 + 2 × 2 5 × 3 2
=
20 − 18
20 + 18 + 12 10 38 + 12 10
=
=
2
2
2 (19 + 6 10)
=
= 19 + 6 10
2
Q.23. Find the values of ‘a’ and ‘b’ in each of the following:
(i)
2+ 3
2– 3
(iii)
(v)
= a +b 3
3
3– 2
= a 3 –b 2
3+4 2
= a–b 3
(ii)
(iv)
(vi)
7 –2
7 +2
= a 7 +b
5+3 2
= a +b 2
5–3 2
4 5+3 2
= a + b 10
3 2 +5 3
3 5–2 2
2+ 3
Ans. (i)
= a+b 3
2− 3
Multiplying numerator and denominator of L.H.S. by (2 + 3) , we get
2 + 3 2 + 3 2 + 3 (2 + 3)2
=
×
=
4−3
2− 3 2− 3 2+ 3
(2)2 + ( 3) 2 + 2 × 2 × 3
{using (a + b)2 = a 2 + 2ab + b 2 }
1
= 4+3+ 4 3 = 7+ 4 3
2+ 3
But
= a + b 3. So, 7 + 4 3 = a + b 3
2− 3
Comparing both sides we get:
a = 7 and b = 4
=
Math Class IX
19
Question Bank
(ii)
7 −2
= a 7 +b
7+2
Multiplying numerator and denominator of L.H.S. by
7 − 2 , we get
7 −2
7 −2
7 − 2 ( 7 − 2)2
=
×
=
7−4
7+2
7 +2
7 −2
( 7)2 + (2)2 − 2 × 2 × 7
=
{using (a − b)2 = a 2 − 2ab + b2 }
3
7 + 4 − 4 7 11 − 4 7
=
=
3
3
11 4 7
7 −2
But
−
= a 7 + b.
= a 7 + b. So,
3
3
7+2
Comparing both sides, we get
−4 7
11
a 7=
and b =
3
3
−4
11
and b =
⇒ a=
3
3
3
(iii)
= a 3 −b 2
3− 2
Multiplying numerator and denominator of L.H.S. by 3 + 2 , we get
3
3
3+ 2
3 3 +3 2
=
×
=
3− 2
3− 2
3 + 2 ( 3)2 − ( 2)2
=
3 3 +3 2 3 3 +3 2
=
3− 2
1
3
= a 3 − b 2. So, 3 3 + 3 2 = a 3 − b 2
3− 2
Comparing both sides, we get
a = 3 and b = −3
Also
Math Class IX
20
Question Bank
(iv)
5+3 2
= a+b 2
5−3 2
Multiplying numerator and denominator of L.H.S. by 5 + 3 2 , we get
5+3 2 5+3 2 5+3 2
(5 + 3 2)2
=
×
=
5 − 3 2 5 − 3 2 5 + 3 2 (5) 2 − (3 2) 2
(5)2 + (3 2) 2 + 2 × 5 × 3 2
=
{using (a + b)2 = a 2 + 2ab + b 2 }
25 − 18
25 + 18 + 30 2 43 + 30 2
=
=
7
7
5+3 2
43 30 2
= a + b 2. So,
+
= a+b 2
Also,
7
7
5−3 2
Comparing both sides, we get :
43
30
a=
and b =
7
7
3+4 2
(v)
= a −b 3
3 2 +5 3
Multiplying numerator and denominator of L.H.S. by 3 2 − 5 3 , we get :
3+4 2
3 + 4 2 3 2 −5 3
=
×
3 2 +5 3 3 2 +5 3 3 2 −5 3
3 2 × 3 − 5 × 3 + 12 × 2 − 20 2 × 3
=
(3 2) 2 − (5 3) 2
3 2 × 3 − 15 + 24 − 20 2 × 3
18 − 75
9 − 17 2 × 3
9 17 6 −3 17 6
=
=− +
=
+
−57
57
57
19
57
−3 17 6
3+4 2
But
+
= a −b 6
= a − b 3. So,
19
57
3 2 +5 3
Comparing both sides, we get :
3
−17
and b =
a=−
19
57
=
Math Class IX
21
Question Bank
(vi)
4 5+3 2
= a + b 10
3 5 −2 2
Multiplying numerator and denominator of L.H.S. by 3 5 + 2 2 , we get
4 5+3 2 4 5+3 2 3 5+2 2
=
×
3 5−2 2 3 5−2 2 3 5+2 2
4 5 ×3 5 + 4 5 ×2 2 + 3 2 ×3 5 + 3 2 × 2 2
=
(3 5)2 − (2 2)2
12 × 5 + 8 10 + 9 10 + 6 × 2
45 − 8
72 + 17 10 72 17 10
=
=
+
37
37
37
72 17 10
4 5+3 2
Also,
+
= a + b 10. So, a + b 10 =
37
37
3 5 −2 2
Comparing both sides, we get :
72
17
a=
and b =
37
37
=
Math Class IX
22
Question Bank