Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Maths Solutions

Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Foundation Mathmatics Solutions. In this article you will learn how to rationalise the  denominator of Irrational numbers. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Maths Solutions

Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Maths Solutions

Board ICSE
Publications Goyal brothers Prakshan
Subject Maths
Class 9th
Chapter-1 Rational and Irrational Numbers
Writer RS Aggrawal
Book Name Foundation
Topics Solution of Exe-1D
Academic Session 2024-2025

How to Rationalise Denominator of Irrational Numbers.

  1. Step 1: Multiply both the denominator and numerator by a suitable conjugate that will remove the radicals from the denominator.
  2. Step 2: We need to make sure that all the surds in the given fraction are in their simplified form.
  3. Step 3: If needed, we can simplify the fraction further

Exercise- 1D

Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Foundation Mathmatics Solutions

Page- 21,22

Rationalise the denominator of each of the following (Q1 to Q11) :

Que-1: 2/√6

Solution- For rationalising denominator, multiply numerator and denominator by √6.
We get, E = 2/√6 × √6/√6
= 2√6/6 = √6/3 Ans.

Que-2: √2/2√3 

Solution- For rationalising denominator, multiply numerator and denominator by 2√3.
We get, E = √2/2√3 × 2√3/2√3
= 2√6/12 = √6/6 Ans.

Que-3: 1/(3+√5)

Solution- For rationalising denominator, multiply numerator and denominator by (3-√5).
We get, E = 1/(3+√5) x (3-√5)/(3-√5)
= (3-√5)/3²-√5²
= 3-√5/9-4 = (3-√5)/4 Ans.

Que-4: 1/(√3-1)

Solution- For rationalising denominator, multiply numerator and denominator by (√3+1).
We get, E = 1/(√3-1) x (√3+1)/(√3+1)
= (√3+1)/√3²-1²
= (√3+1)/3-1 = (√3+1)/2 Ans.

Que-5: 1/(4+2√3)

Solution-  For rationalising denominator, multiply numerator and denominator by (4-2√3).
We get, E = 1/(4+2√3) x (4-2√3)/(4-2√3)
= (4-2√3)/4²-(2√3)²
= (4-2√3)/16-12
= (4-2√3)/4 = (2-√3)/2 Ans.

Que-6: 1/(√6-√3)

Solution- For rationalising denominator, multiply numerator and denominator by (√6+√3).
We get,  E = 1/(√6-√3) x (√6+√3)/(√6+√3)
= (√6+√3)/√6²+√3²
= (√6+√3)/6-3 = (√6+√3)/3 Ans.

Que-7: √3-1/√3+1

Solution- For rationalising denominator, multiply numerator and denominator by (√3-1).
We get,  E = (√3-1)/(√3+1) x (√3-1)/(√3-1)
= 3-√3-√3+1/√3²-1²
= 4-2√3/3-1 = (2-√3) Ans.

Que-8: (3-2√2)/(3+2√2)

Solution- For rationalising denominator, multiply numerator and denominator by (3-2√2).
We get,  E = (3-2√2)/(3+2√2) x (3-2√2)/(3-2√2)
= (9-6√2-6√2+8)/3²-(2√2)²
= 17-12√2/9-8 = 17-12√2 Ans.

Que-9: 1/(2√5-√3)

Solution- For rationalising denominator, multiply numerator and denominator by (2√5+√3).
We get,  E = 1/(2√5-√3) x (2√5+√3)/(2√5+√3)
= (2√5+√3)/(2√5)²-(√3)²
= (2√5+√3)/20-3 = (2√5+√3)/17 Ans.

Que-10: 1/(1+√5+√3)

Solution- For rationalising denominator, multiply numerator and denominator by
[(1+√5)-√3].  We get,
E = 1/(1+√5+√3) x [(1+√5)-√3]/[(1+√5)-√3]
= [(1+√5)-√3]/(1+√5)²-(√3)²
= [(1+√5)-√3]/(1+5+2√5-3)
= [(1+√5)-√3]/(3+2√5)
Again, rationalise the denominator 3+2√5 by 3-2√5
We get,  = [(1+√5)-√3]/(3+2√5) x (3-2√5)/(3-2√5)
= [3-2√5+3√5-10-3√3+2√15]/(9-20)
= [-7-3√3+√5+2√15]/-11
= [7+3√3-√5-2√15]/11 Ans.

Que-11: √2/(√2+√3-√5)

Solution- For rationalising denominator, multiply numerator and denominator by
[(√2+√3)+√5].  We get,
E = √2/(√2+√3-√5) x [(√2+√3)+√5]/[(√2+√3)+√5]
= √2[(√2+√3)+√5]/(√2+√3)²-(√5)²
= √2[(√2+√3)+√5]/(2+3+2√6-5)
= √2[(√2+√3)+√5]/(2√6)
Again, rationalise the denominator 2√6 by 2√6
We get,  = √2[(√2+√3)+√5]/2√6 x 2√6/2√6
= 2√6[2+√6+√10]/4×6
= 4√6+12+4√15/24
= 4(√6+3+√15)/24
= (√6+3+√15)/6 Ans.

Que-12: If (√3+1)/(√3-1) = a+b√3, find the values of a and b.

Solution-√3−1√3+1
=√3−1√3+1×√3−1√3−1
=3+1−2√33−1
=4−2√32
=2(2−√3)2
=2−√3
⇒a=2, b=1 Ans.

Que-13: If (3+√2)/(3-√2) = a+b√2, find the value of a and b.

Solution- First rationalise : (3+√2)/(3-√2)
For this multiply both denominator and numerator with conjugate of (3-√2) that is (3+√2)
⇒ (3+√2)/(3-√2) × (3+√2)/(3+√2)
⇒ (3+√2)²/(3-√2) (3+√2)            { (a+b) (a-b) = a²- b²}
⇒ (9 + 2 + 6√2)/ 9 – 2
⇒ (11 + 6√2) / 7
compare with a + b√2
so, a = 11/7 and b=6/7
Hence,  the value of a is 11/7 and b is 6/7.

Que-14: If (5-√6)/(5+√6) = a-b√6, find the value of a and b.

Solution- (5-√6)/(5+√6)= a-b√6
multiply by conjugate on both numerator and denominator in the LHS
(5-√6)(5-√6)/(5+√6)(5-√6) = a-b√6
(5-√6)²/(5)²-(√6)² = a-b√6
(25+6-10√6)/(25-6) = a-b√6
(31-10√6)/19 = a-b√6
31/19 -10√6/19 = a-b√6
therefore
a = 31/19 and b = 10/19 Ans.

Que-15: If (5+2√3)/(7+4√3) = a-b√3, find the value of a and b.

Solution-  (5 + 2√3) / (7 +4 √3) = a + b√3
By rationalization, we get
(5 + 2√3)(7 – 4√3) / (7 +4 √3) (7 – 4√3) = a + b√3
35 – 20√3 + 14√3 – 24 = a + b√3
11 – 6√3 = a + b√3
On equating the above equation, we get
a = 11 and b = 6 Ans.

Que-16: Simplify : (√5+√3)/(√5-√3) + (√5-√3) / (√5+√3)

Solution-(√5+√3)/(√5−√3) + (√5−√3)/(√5+√3)
= [(√5+√3)²+(√5−√3)²]/(√5−√3)(√5+√3)
= 2[(√5)²+(√3)²]/(√5)²−(√3)²                           {∵(a+b)2+(a−b)2=2(a2+b2)}
= [2(5+3)]/(5−3)
=(2×8)/2 = 8 Ans.

Que-17: Simplify : (7+3√5)/(3+√5) – (7-3√5)/(3-√5)

Solution- On rationalising we get
(7+3√5)/(3+√5) x (3-√5)/(3-√5) – (7-3√5)/(3-√5) x (3+√5)
Using the identity,
(a+b)(a-b) = a²-b²
= [7×3-7x√5+3√5×3-3√5x√5]/(3)²-(√5)²
= [7×3-7√5+9√5-3×5]/9-5
= [21-7√5+9√5-15]/4 – [21+7√5+9√5-15]/4
= [6+2√5]/4 – [6-2√5]/4
= [6+2√5-6+2√5]/4
= 4√5/4 = √5 Ans.

Que-18: Show that : 1/(3-√8) + 1/(√7-√6) + 1/(√5-2) – 1/(√8-√7)
– 1/(√6-√5) = 5

Solution-  1/(3-√8) + 1/(√7-√6) + 1/(√5-2) – 1/(√8-√7) – 1/(√6-√5) = 5
Rationalise all the denominators,
[(3+√8)/1] − [(√8+√7)/1] + [(√7+√6)/1] − [(√6+√5)/1] + [(√5+2)/1]
= 3 + √8 − √8 − √7 + √7 + √6 − √6 − √5 + √5 + 2 = 5.

Que-19: If x = (3+√8), find the values of (x²+1/x²).

Solution- Let us find the value of 1/x using the vale given in equation 1:
x = 3 + √8
then, 1/x = 1/(3 + √8) (equation 2)
Now, we have to rationalize the denominator of the value obtained in equation 2.
So, for rationalization we will multiply the numerator and denominator of 1/(3 + √8) with (3 – √8).
1/x = 1 x (3 – √8) / (3 + √8) x (3 – √8)
1/x = (3 – √8) / (3)2 + 3√8 – 3√8 – (√8)2
1/x = (3 – √8) / (3)2 – (√8)2
1/x = (3 – √8) / (9 – 8)
1/x = (3 – √8) / 1
1/x = (3 – √8)
Now, we have the value of x = (3 + √8) and the value of 1/x = (3 – √8)
So, the value of x2 + 1/x2 = (3 + √8)2 + (3 – √8)2 (equation 3)
Apply (a+b)2 and (a-b)2 formula in (equation 3)
x2 + 1/x2 = (3)2 + (√8)2 + 2 x 3 x √8 + (3)2 + (√8)2 – 2 x 3 x √8
x2 + 1/x2 = (3)2 + (√8)2 + (3)2 + (√8)2
x2 + 1/x2 = 9 + 8 + 9 + 8
x2 + 1/x2 = 34 Ans.

Que-20: If x = (4-√15), find the values of (x²+1/x²).

Solution-  ⇒ x2 = (4 + √15)2
⇒ 16 + 15 + 8√15 = 31 + 8√15
And
⇒ 1/x2 = 1/(31 + 8√15)
⇒ (31 – 8√15)/(961 – 960) = 31 – 8√15
Now, put the values in the equation:
⇒ x2 + (1/x2)
⇒ 31 + 8√15 + 31 – 8√15 = 62 Ans.

–: End of Rational and Irrational Numbers Class 9 RS Aggarwal Exe-1D Goyal Brothers ICSE Maths : —

Return to :- RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)

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