ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.4
Exercise 1.4
Question 1
(i) =√45−3√20+4√5
=√9×5−3√4×5+4√5
=√9√5−3√4√5+4√5
=3√5−3×2√5+4√5
=3√5−6√5+4√5
=(3−6+4)√5
=(7−6)√5
=√5
(ii) =3√3+2√27+7√3
=3√3+2√9×3+7√3
=3√3+2×√9√3+7√3
=3√3+2×3√3+7√3
=3√3+6√3+7√3
=(3+6)√3+7√3
=9√3+7√3
Multiplying and Dividing by "√3"
=9√3+7√3×√3√3
=9√3+7√33
By cross multiplying
=3×9√3+7√33
=27√3+7√33
=(27+7)√33
=34√33
(iii) 6√5×2√5
6×2×√5⋅√5
=12×(√5)2
=12×5
=60
(iv) 8√15÷2√3
8√152√3=8√5√32√3
=4√5
(v) √248+√549
Sol :
√6×48+√9×69
=√6⋅√48+√9⋅√69
=2√68+3√69
=1⋅√64+1⋅√63
=√6[14+13]
=√6[3+412] [∵LCM of 4 and 3 is 12]
=7√612
(vi) 3√8+1√2
Sol :
3√2×4+1√2
=3√2⋅√4+1√2
=32√2+1√2
=1√2(32+1)
=1√2(3+22) [∵LCM of 2 and 1 is 2]
=1√2(52)
=52√2
Multiply and Divide by '√2'
=52√2×√2√2
=5√22⋅√2×√2
=5√22(√2)2
=5√22⋅2
=5√24
Question 2
Sol :
(i) (5+√7)(2+√5)
Sol :
=5×2+5√5+2√7+√7⋅√5
=10+5√5+2√7+√7×5
=10+5√5+2√7+√35
(ii) (5+√5)(5−√5)
=(5)2−(√5)2
=25−5
=20
(iii) (√5+√2)2
=(√5)2+(√2)2+2⋅√5⋅√2
=5+2+2√5×2
=5+2+2√10
=7+2√10
(iv) (√3−√7)2
=(√3)2+(√7⋅)2−2⋅√3⋅√7
=3+7−2√3×7
=10−2√21
(v) (√2+√3)(√5+√7)
=√2⋅√5+√2⋅√7+√3⋅√5+√3⋅√7
=√2×5+√2×7+√3×5+√3×7
=√10+√14+√15+√21
(vi) (4+√5)(√3−√7)
=4√3−4√7+√5⋅√3−√5⋅√7
=4√3−4√7+√5×3−√5×7
=4√3−4√7+√15−√35
Question 3
Sol :
=√4×2+√25×2+√36×2+√49×2
=√4⋅√2+√25⋅√2+√36⋅√2+√49⋅√2
=2√2+5√2+6√2+7√2
=(2+5+6+7)√2
=20×√2
=20×1⋅414
=28.28
(ii) 3√32−2√50+4√128−20√18
=3√16×2−2√25×2+4√64×2−20√9×2
=3×4√2−2×5√2+4×8√2−20×3√2
=12√2−10√2+32√2−60√2
=(12−10+32−60)√2
=−26×√2
=−26×1.414
=-36.764
Question 4
Sol :
(i) √27+√75+√108−√243
√9×3+√25×3+√36×3−√81×3
=3√3+5√3+6√3−9√3
=(3+5+6−9)√3
=(14−9)√3
=5×√3
=5×1.732
=8.66
(ii) 5√12−3√48+6√75+7√108
Sol :
=5√4×3−3√16×3+6√25×3+7√36×3
=5×2√3−3×4√3+6×5√3+7×6√3
=10√3−12√3+30√3+42√3
=(10−12+30+42)√3
=(82−12)√3
=70×√3
=70×1.732
=121.24
Question 5
(i) √49,−370,√725,√165
Sol :
√49=√4√9
=23
It is in the form of pqand p, q are integers
Therefore √49 is a rational number
⇒−370 is a rational number
⇒√725=√7√25
=√75
Since √7 is not an Integer
Therefore, √725 is an Irrational number
√165=√16√5
=4√5. Since √5 is not an Integer
Therefore, √165 is an Irrational number
(ii) −√249,3200,√253,−√4916
Sol :
−√249=−√2√49
=−√27. Since √2 is not an Integer
Therefore, −√249 is an Irrational number
3200. It is in the form of pq and p, q are integers. So, it is a rational number
√253=√25√3
=5√3⋅ since √3 is not an integer
Therefore, √253 is an irrational number
−√4916=−√49√16
=−74
Therefore, −√4916 is a rational number
Question 6
(i) −3√2
Sol :
Since √2 is an lrrational number
Therefore, −3√2 will change into non-terminating , non-recurring decimal
(ii) √25681
Sol :
√256√81=169
=1.77777¯7
Therefore, √25681 will not change into non-terminating, non - recurring decimal
(iii) √27×16
Sol :
√27×√16=√9×3×4
=4×3√3
=12√3
∴√3 is an Irrational number
Therefore, √27×16 will change into non-terminating, non-recurring decimal
(iv) √536
Sol :
√5√36=√56
∵√5 is an Irrational number
Therefore, √536 will change into non-terminating ,non - recurring decimal
(iv) −273√5
Sol :
Since 3√5 is an lrrational number
Therefore, −273√5 is also an Irrational number
(Note: Product of rational and Irrational number is irrational)
(v) (2−√3)(2+√3)
Sol :
=2×2+2√3−2√3−(√3)2
=4-3=1
Therefore (2−√3)(2+√3) is a rational number
(vi) (3+√5)2
Sol :
=(3)2+(√5)2+2×3×√5
=9+5+6√5
=14+6√5
Since √5 is an Irrational number
(3+√5)2 is also an Irrational number
(vii) (25√7)2
Sol :
(25)2⋅(√7)2
=425×7
=2825
Therefore, (25√7)2 is a rational number
(viii) (3−√6)2
Sol :
(3)2+(√6)2−2×3×√6
=9+6−6√6
=15−6√6
Since √6 is an Irrational number (3−√6)2 is also an irrational . number
Question 7
Question 8
(i) 3√2
Sol :
Suppose that 3√2=pq, where p, q are integers , q>0, p and q have no common factors (except 1)
2=[pq]3
p3=2q3...(1)
As 2 divides 2q3⇒2 divides p3
2 divides p
Let p=2k, where k is an integer
Substituting this value of 'p' in (1), we get
(2k)3=2q3
8k3=2q3
4k3=q3
As 2 divides 4k3⇒2 divides q3
2 divides q
Thus p and q have a common factor '2'
This contradicts that p and q have no common factor (except 1)
Therefore , 3√2 is an irrational number
(ii) 3√3
Sol :
Suppose that 3√3=pq, where p, q are integers, q>0 , p and q have no common factors (except 1)
3=(pq)3
p3=3q3...(1)
As 3 divides 3q3⇒3 divides p3
3 divides p
Let p=3k, Where k is an Integer
Substituting this value of 'p' in 0 , We get
(3k)3=3q3
27k3=3q3
9k3=q3
As 3 divides 9K3⇒3 divides q3
3 divide q
Thus p and q have a common factor '3'
This contradicts that p and q have no common factor (except 1)
Therefore 3√3 is an Irrational number
(iii) 4√5
Sol :
Suppose that 4√5=pq, where p, q are integers , q>0, p and q have no common factors (except 1)
5=(pq)4
p4=5q4...(1)
As 5 divides 5q4⇒5 divides p4
5 divides p
Let p=5k , where k is an integer
Substituting this value of 'p' in (1) , we get
(5k)4=5q4
625k4=5q4
125k4=q4
As 5 divides 125 k4⇒5 divides q4
5 divides q
Thus p and q have a common factor "5"
This contradicts that p and q have no common factors (except 1)
Therefore, 4√5 is an Irrational number.
Question 9
(i) 2√3,3√2,−√7,√15
Sol :
√4×3=√12
3√2=√92
√4⋅5
∴√12,√4⋅5,−√7,√15
−√7<√4⋅5<√12<√15
The greatest real number is √15
The smallest real number is −√7
(ii) −3√2,9√5,−4,43√5,32√3
Sol :
−3√2=−√9×2
=−√18
9√5=√815
=√16.2
−4=−√16
43√5=√16×59
=√809
=√8.89
32√3=√9×34
=√274
=√6.75
∴−√18,√16.2,√8.89,−√16,√6.75
−3√2<−4<32√3<43√5<9√5
The greatest real number is 9√5
The smallest real number is −3√2
Question 10
(i) 3√2,2√3,√15,4
Sol :
White all the numbers as square roots under one radical
3√2=√9×√2=√18
2√3=√4×√3=√12
√15=√15
4=√16
Since 12<15<16<18
⇒√12<√15<√16<√18
⇒2√3<√15<4<3√2
Hence, the given numbers in ascending orders are
2√3,√15,4,3√2
(ii) 3√2,2√8,4,√50,4√3
Sol :
Write an the numbers as square roots under one radical
3√2=√9×√2=√18
2√8=√4×√8=√32
4=√16
√50=√50
4√3=√16×√3=√48
Since 16<18<32<48<50
⇒√16<√18<√32<√48<√50
⇒4<3√2<2√8<4√3<√50
Question 11
(i) 9√2,32√5,4√3,3√65
Sol :
Write all the numbers as square roots under one radical
9√2=√812=√40.5
32√5=√94×√5=√454=√11⋅25
4√3=√16×√3=√48
3√65=√9×√65=√545=√10.8
Since 48>40.5>11.25>10.8
√48>√40.5>√11.25>√10.8
4√3>9√2>32√5>3√65
Hence, the given numbers in descending orders are 4√3,9√2,32√5,3√65
(ii) 5√3,73√2,−√3,3√5,2√7
Sol :
Write all the numbers as Square roots under one radical
5√3=√253=√8.33
73√2=√499×√2
=√989=√10.89
−√3=−√3
3√5=√9×√5=√45
2√7=√4×√7=√28
Since 45>28>10.89>8.33 >-3
√45×√28>√10.89>√8.33>−√3
3√5>2√7>73√2>5√3>−√3
Hence, the given numbers in descending orders are 3√5,2√7,73√2,5√3,−√3
Question 12
(i) 3√2,√3,6√5
Sol :
L.C.M of 2,3 and 6 is 6
3√2=213=(22)16=(4)16
√3=312=(33)16=(27)16
6√5=516=(51)16=(5)16
Since 4<5<27
(4)16<(5)16<(27)16
3√2<6√5<√3
Hence, the given numbers in ascending orders are 3√2,6√5,√3
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