Expansions Class 9 RS Aggarwal Exe-3A Goyal Brothers ICSE Maths Solutions

Expansions Class 9 RS Aggarwal Exe-3A Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn How to Expand (a+b)^2 and  (a-b)^2 Using Formula  easily. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Expansions Class 9 RS Aggarwal Exe-3A Goyal Brothers ICSE Maths Solutions

Board	ICSE
Publications	Goyal brothers Prakshan
Subject	Maths
Class	9th
Chapter-3	Expansion
Writer	RS Aggrawal
Book Name	Foundation
Topics	Special Products
Academic Session	2024-2025

How to Expand (a+b)^2 and  (a-b)^2 Using Formula

Expand the (a+b)^2 and  (a-b)^2  as given below. apply these two expansion formula to expand the expression.

1. (a + b)² = a² + 2ab + b²
2. (a – b)² = a² – 2ab + b²

Using (a + b)² formula: a² + b² = (a + b)² – 2ab. Using (a – b)² formula: a² + b² = (a – b)² + 2ab

Exercise- 3A

Expansions Class 9 RS Aggarwal Exe-3A Goyal Brothers ICSE Foundation Maths Solutions

Page- 48,49

Using the standard formulae, expand each of the following (Q1 to 13) :

Que-1: (i) (4a+9)²   (ii) (3x+10y)²    (iii) (√2m+√3n)²

Solution- (a+b)² = a²+b²+2ab
(i) (4a +9)² = (4a)² + 2 (4a) (9) + 9²
= 16a² + 72a + 81
(ii) (3x +10y)² = (3x)² + 2 (3x) (10y) + (10y)²
= 9x² + 60xy + 100y²
(iii) (√2m +√3n)² = (√2m)² + 2 (√2m) (√3n) + (√3n)²
= 2m² + 2√6mn + 3n²

Que-2: (i) (2a²+3b)²   (ii) (3x²y+z)²   (iii) (2x+1/3x)²

Solution- (a + b)² = a² + b² + 2ab
(i) (2a² + 3b)²
(2a²)² + (3b)² + 2(2a²)(3b)
4a⁴ + 9b² + 12a² b
(ii) (3x² y + z)²
(3x² y)² + (z)² + 2(3x² y)(z)
9x⁴ y2 + z² + 6 x²yz
(iii) (2x + 1/3x)²
(2x)² + (1/3x)² + 2(2x)(1/3x)
4x² + 1/9x² + 4/3x²

Que-3: (i) (2x/5+5y/6)²    (ii) (x/3+6/x)²    (iii) (6+5/x)²

Solution- (a-b)² = a² + b² + 2ab
(i) (2x/5+5y/6)²
= (2x/5)² + (5y/6)² + 2(2x/5)(5y/6)
= 4x²/25 + 25y²/36 + 2xy/3
(ii) (x/3+6/x)²
= (x/3)² + (6/x)² + 2(x/3)(6/x)
= x²/9 + 36/x² + 4
(iii) (6+5/x)²
= (6)² + (5/x)² + 2(6)(5/x)
= 36 + 25/x² + 60/x.

Que-4: (i) (5x-3y)²   (ii) (3a-7b)²    (iii) (1x/2-3y/2)²

Solution- (a-b)² = a² + b² – 2ab
(i) (5x-3y)²
= (5x)² + (3y)² – 2(5x)(3y)
= 25x² + 9y² – 30xy
(ii) (3a-7b)²
= (3a)² + (7b)² – 2(3a)(7b)
= 9a² + 49b² – 42ab
(iii) (1x/2-3y/2)²
= (1x/2)² + (3y/2)² – 2(1x/2)(3y/2)
= x²/4 + 9y²/4 – 3xy/2

Que-5: (i) (a²-b/2)²   (ii) (3a/2b – 2b/3a)²    (iii) (5x-2/3x)²

Solution- (a-b)² = a² + b² – 2ab
(a²-b/2)²
= (a²)² + (b/2)² – 2(a)(b/2)
= a^4 + b²/4 – a²b
(ii) (3a/2b-2b/3a)²
= (3a/2b)² + (2b/3a)² – 2(3a/2b)(2b/3a)
= 9a²/4b² + 4b²/9a² – 2
(iii) (5x-2/3x)²
= (5x)² + (2/3x)² – 2(5x)(2/3x)
= 25x² + 4/9x² – 20/3.

Que-6: (i) (a+2b+3c)²   (ii) (3x+5y-2z)²    (iii) (2x-3y+7z)²

Solution- (i) (𝑎+2𝑏+3𝑐)²
According to the equation ,
(𝑎+𝑏+𝑐)² = 𝑎²+𝑏²+𝑐²+2𝑎𝑏+2𝑏𝑐+2𝑐𝑎
So we get,
Using the formula we can it write it as
(𝑎+2𝑏+3𝑐)² = 𝑎²+(2𝑏)²+(3𝑐)²+2𝑎(2𝑏)+(2(2𝑏)(3𝑐)+2(3𝑐)𝑎
On further calculation we get,
(𝑎+2𝑏+3𝑐)² = 𝑎²+4𝑏²+9𝑐²+4𝑎𝑏+12𝑏𝑐+6𝑎𝑐

(ii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(3x+5y-2z)² = (3x)²+(5y)²+(-2z)²+(2×3x×5y)+{2×5y×(-2z)}+{2×(-2z)+3x}
= 9x²+25y²+4z²+30xy-20yz-12zx.

(iii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(2x-3y+7z)² = (2x)²+(-3y)²+(7z)²+(2.2x.(-3y))+(2.(-3y).7z)+(2.7z.2x)
= 4x²+9y²+49z²-12xy-42yz+28xz

Que-7: (i) (6-2y+4z)²   (ii) (4x-3y+z)²    (iii) (7-2x-3y)²

Solution- (i) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(6-2y+4z)² = (6)²+(-2y)²+(4z)²+{2.6.(-2y)}+{2.(-2y).4z}+{2.6.4z}
= 36+4y²+16z²-24y-16z+48z

(ii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(4x-3y+z)² = (4x)²+(-3y)²+(z)²+{2.4x.(-3y)}+{2.(-3y).z}+{2.z.4x}
= 16x²+9y²+z²-24xy-6yz+8xz

(iii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(7-2x-3y)² = (7)²+(-2x)²+(-3y)²+{2.7.(-2x)}+{2.(-2x).(-3y)}+{2.7.(-3y)}
= 49+4x²+9y²-28x+12xy-42y.

Que-8: (i) (a/2 + b/3 + c/4)²   (ii) (2x/3+3/2y-2)²    (iii) (2x+3/x-1)²

Solution- (i) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(a/2+b/3+c/4)² = (a/2)²+(b/3)²+(c/4)²+{2.(a/2).(b/3)}+{2.(b/3).(c/4)}+{2.(c/4).(a/2)}
= a²/4 + b²/9 + c²/16 + ab/3 + bc/6 + ac/4

(ii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(2x/3+3/2y-2)² = (2x/3)²+(3/2y)²+(-2)²+{2.(2x/3).(3/2y)}+{2.(3/2y).(-2)}+{2.(-2).(2x/3)}
= 4x²/9 + 9/4y² + 4 + 2x/y – 6/y – 8x/3

(iii) (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(2x+3/x-1)² = (2x)²+(3/x)²+(-1)²+{2.(2x).(3/x)}+{2.{3/x}.(-1)}+{2.(-1).(2x)}
= 4x² + 9/x² + 1 + 12 – 6/x – 4x
= 4x² + 9/x² + 13 – 6/x – 4x.

Que-9: (i) (x+7)(x+4)   (ii) (a+13)(a-8)    (iii) (y-6)(y-4)

Solution- (i) (x+7)(x+4)
= x²+4x+7x+28
= x²+11x+28
(ii) (a+13)(a-8)
= a²-8a+13a-104
= a²+5a-104
(iii) (y-6)(y-4)
= y²-4y-6y+24
= y²-10y+24.

Que-10: (i) (9+2x)(9-3x)   (ii) (5x-4y)(5x+3y)    (iii) (3-7a)(3+4a)

Solution- (i) (9+2x)(9-3x)
= 81-27x+18x-6x²
= 81-9x-6x²
(ii) (5x-4y)(5x+3y)
= 25x²+15xy-20xy-12y²
= 25x²-5xy-12y²
(iii) (3-7a)(3+4a)
= 9+12a-21a-28a²
= 9-9a-28a².

Que-11: (i) (3a+2b)(3a-2b)    (ii) (5x+1/5x)(5x-1/5x) (iii)  (2x²+3/x²) (2x²-3/x²)

Solution- (i) (3a+2b)(3a-2b)
= 9a²-6ab+6ab-4b²
= 9a² – 4b²
(ii) (5x+1/5x)(5x-1/5x)
[Using identity, (a + b)(a – b) = a² – b²]
a = 5x  and  b = 1/5x
= (5x)² – (1/5x)²
= 25x² – 1/25x²
(iii) (2x²+3/x²)(2x²-3/x²)
[Using identity, (a + b)(a – b) = a² – b²]
a = 2x²   and   b = 3/x²
= (2x²)² – (3/x²)²
= 4x^4 – 9/x^4.

Que-12: (i) (2-x)(2+x)(4+x²)    (ii) (x+y)(x-y)(x²+y²)

Solution- (i) (2-x)(2+x)(4+x²)
[Using identity, (a + b)(a – b) = a² – b²]
= [(2)² – (x)²] (4+x²)
= (4-x²)(4+x²)
[Using identity, (a + b)(a – b) = a² – b²]
= (4)² – (x²)²
= 16 – x^4
(ii) (x+y)(x-y)(x²+y²)
[Using identity, (a + b)(a – b) = a² – b²]
= [(x)² – (y)²] (x²+y²)
= (x²-y²)(x²+y²)
[Using identity, (a + b)(a – b) = a² – b²]
= (x²)² – (y²)²
= x^4 – y^4.

Que-13: (i) (x-2)(x-3)(x+4)   (ii) (x-5)(2x-1)(2x+3)

Solution- (i) (x-2)(x-3)(x+4)
= [x²-3x-2x+6](x+4)
= [x²-5x+6](x+4)
= [x³-5x²+6x+4x²-20x+24]
= [x³-x²-14x+24]

(ii) (x-5)(2x-1)(2x+3)
= [2x²-x-10x+5](2x+3)
= [2x²-11x+5](2x+3)
= [4x³-22x²+10x+6x²-33x+15]
= [4x³-16x²-23x+15]

Que-14: Simplify :

(i) (a+b)² + (a-b)²   (ii) (a+b)² – (a-b)²   (iii) (x+1/x)² + (x-1/x)²
(iv) (x+1/x)² – (x-1/x)²   (v) (a/2b+2b/a)² – (2b/a-a/2b)²
(vi) (3x-1/3x)² – (3x+1/3x)(3x-1/3x)   (vii) (5a+3b)² – (5a-3b)² – 60ab
(viii) (3x+1)² – (3x+2)(3x-1)

Solution- (i) (a+b)² + (a-b)²
= (a²+b²+2ab) + (a²+b²-2ab)
= [a²+b²+2ab+a²+b²-2ab]
= 2a²+2b²
= 2(a²+b²)

(ii) (a+b)² – (a-b)²
[Using identity, (a + b)(a – b) = a² – b²]
= [a+b+a-b] [a+b-a+b]
= (2a)(2b)
= 4ab

(iii) (x+1/x)² + (x-1/x)²
= [x²+1/x²+2.x.1/x] + [x²+1/x²-2.x.1/x]
= [x²+1/x²+2] + [x²+1/x²-2]
= [x²+1/x²+2+x²+1/x²-2]
= 2x²+2/x²
= 2(x²+1/x²)

(iv) (x+1/x)² – (x-1/x)²
[Using identity, (a + b)(a – b) = a² – b²]
[x+1/x+x-1/x] [x+1/x-x+1/x]
= (2x)(2/x)
= 4

(v) (a/2b+2b/a)² – (2b/a-a/2b)²
[Using identity, (a + b)(a – b) = a² – b²]
= [a/2b+2b/a+2b/a-a/2b] [a/2b+2b/a-2b/a+a/2b]
= (4b/a)(2a/2b)
= 4

(vi) (3x-1/3x)² – (3x+1/3x)(3x-1/3x)
[Using identity, (a + b)(a – b) = a² – b²]
= (3x-1/3x)² – [(3x)² – (1/3x)²]
= [(3x)²+(1/3x)²-2.3x.1/3x] – (9x²-1/9x²)
= [9x²+1/9x²-2] – (9x²-1/9x²)
= [9x²+1/9x²-2-9x²+1/9x²]
= 2/9x²-2
= 2(1/9x²-1)

(vii) (5a+3b)² – (5a-3b)² – 60ab
= [(5a)²+(3b)²+2.5a.3b] – [(5a)²+(3b)²-2.5a.3b] – 60ab
= [25a²+9b²+30ab] – [25a²+9b²-30ab] – 60ab
= 25a²+9b²+30ab-25a²-9b²+30ab-60ab
= 60ab – 60ab = 0.

(viii) (3x+1)² – (3x+2)(3x-1)
= [(3x)²+(1)²+2.1.3x] – (9x²-3x+6x-2)
= (9x²+1+6x) – (9x²+3x-2)
= 9x²+1+6x-9x²-3x+2
= 3x+3
= 3(x+1).

Que-15: (i) If a+b = 7 and ab = 10, find the value of (a-b) , (ii) If x-y = 5 and xy = 24, find the value of (x+y)

Solution- (i) We know that,
( a + b )² = a² + 2ab + b²
and
( a – b )² = a² – 2ab + b²
Rewrite the above equation, we have
( a – b )² = a²  + b² – 2ab + 4ab
= ( a + b )² – 4ab              …(1)
Given that a + b = 7; ab = 10
Substitute the values of ( a + b ) and (ab)
in equation (1), we have
( a – b )² = (7)² – 4(10)
= 49 – 40 = 9
⇒ a – b = ±√9
⇒ a – b = ±3 Ans.

(ii) (x + y)² = (x – y)² + 4xy
= (x + y)² = (5)² + 4 × 24
= (x + y)² = 25 + 96
= (x + y)² = 121
= (x + y) = √121
= (x + y) = 11 Ans.

Que-16: If (3a+4b) = 16 and ab = 4, find the value of (9a²+16b²).

Solution- Given:  (3a+4b)=16 … (i) and ab = 4 … (ii)
Squaring (i), we get
(3a+4b)² = 16²
9a² + 16b² + 2 x 3a x 4b = 256
9a² + 16b² + 24ab = 256
9a² + 16b² = 256 – 24ab
9a² + 16b² = 256 – 24(4) … Using (ii)
9a² + 16b² = 256 – 96
9a² + 16b² = 160 Ans.

Que-17: If (a+b) = 2 and (a-b) = 10, find the values of (i) (a²+b²)  (ii) ab.

Solution- (i) We know that
(a + b)² = a² + b² + 2ab
Given : a + b = 2
ab = – 24
⇒ (2)² = a² + b² + 2(-24)
⇒ 4 = a² + b² – 48
⇒ 4 + 48 = a² + b²
⇒ 52 = a² + b²
⇒ a² + b² = 52 Ans.

(ii) We know that,
(a + b)² = (a – b)² + 4ab
Given : a + b = 2
a – b = 10
⇒ (2)² = (10)² + 4ab
⇒ 4 = 100 + 4ab
⇒ 4 – 100 = 4ab
⇒ – 96 = 4ab
⇒ – 96/4 = ab
⇒ – 24 = ab
⇒ ab = – 24 Ans.

Que-18: If (a-b) = 0.9 and ab = 0.36, find the values of (i) (a+b) (ii) (a²-b²)

Solution- (i) We know that,
( a – b )² = a² – 2ab + b²
and
( a + b )² = a² + 2ab + b²
Rewrite the above equation, we have
( a + b )² = a²  + b² – 2ab + 4ab
= ( a – b )² + 4ab              …(1)
Given that a – b = 0.9 ; ab = 0.36
Substitute the values of ( a – b ) and (ab)
in equation (1), we have
( a + b )² = ( 0.9 )² + 4( 0.36 )
= 0.81 + 1.44 = 2.25
⇒ a + b = ±√2.25
⇒ a + b = ±1.5 Ans.

(ii) We know that,
a² – b² = ( a + b )( a – b )             ….(3)
From equation (2) we have,
a + b = ±1.5
Thus equation (3) becomes,
a² – b² = (±1.5)(0.9)                [ given a – b = 0.9 ]
⇒ a² – b² = ±1.35 Ans.

Que-19: If (x+1/x) = 5, find the values of (i) (x²+1/x²)  (ii) (x^4 + 1/x^4)

Solution- (i) x+1/x = 5
Squaring both sides
(x+1/x)² = (5)²
⇒ x²+1/x²+2×x×1/x = 25
⇒ x²+1/x²+2 = 25
⇒ x²+1/x² = 25-2 = 23
∴x²+1/x² = 23 Ans.

(ii) (x²+1/x²)² = (23)²
⇒ x^4 + 1/x^4 + 2×x²×1/x² = 529
⇒ x^4 + 1/x^4 + 2 = 729
⇒ x^4 + 1/x^4 = 529−2 = 527
∴ x^4 + 1/x^4 = 527 Ans.

Que-20: If (x-1/x) = 4, find the values of (i) (x²+1/x²)  (ii) (x^4 + 1/x^4)

Solution-x-1/x = 4
Squaring both sides,
(i) (x-1/x)² = (4)²
⇒ x² + 1/x² – 2×x+1/x = 16
⇒ x²+1/x² – 2 = 16
⇒ x²+1/x² = 16+2 = 18
∴ x²+1/x² = 18 Ans.

(ii) Again squaring both sides,
(x²+1/x²) = (18)²
⇒ (x²)² + (1/x²)² + 2×x²×1/x² = 324
⇒ x^4 + 1/x^4 + 2 = 324
⇒ x^4 + 1/x^4 = 324−2 = 322
∴ x^4+1/x^4 = 322 Ans.

Que-21: If x-2 = 1/3x, find the values of (i) (x²+1/9x²)  (ii) (x^4+1/81x^4)

Solution-  (i) x-2 = 1/3x
x-1/3x = 2
Squaring on both sides, we get
(x-1/3x)² = (2)²
x² + 1/9x² – 2/3 = 4
x² + 1/9x² = 4 + 2/3
x² + 1/9x² = (12+2)/3
x² + 1/9x² = 14/3 Ans.

(ii) x² + 1/9x² = 14/3.
Squaring on both sides, we get
(x² + 1/9x²)² = (14/3)²
x^4 + 1/81x^4 + 2/9 = 196/9
x^4 + 1/81x^4 = 196/9 – 2/9
x^4 + 1/81x^4 = (196-2)/9
x^4 + 1/81x^4 = 194/9 Ans.

Que-22: If (x+1/x) = 6, find the values of (i) (x-1/x)  (ii) (x²-1/x²).

Solution- (i) x + 1/x = 6
Using algebraic identity (a – b)² = (a + b)² – 4ab
(x – 1/x)² = (x + 1/x)² – 4(x)(1/x)
(x – 1/x)² = 6² – 4(x)(1/x)
(x – 1/x)² = 36 – 4 = 32
x – 1/x = ± √32 = ± 4√2 Ans.

(ii) Now, x² – 1/x² = x² + ( 1/x )²
= ( x + 1/x )( x – 1/x )
Since a² – b² = (a + b)(a – b)
= 6( ± 4√2 )
= ± 24√2 Ans.

Que-23: If (x-1/x) = 8, find the values of (i) (x+1/x)  (ii) (x²-1/x²).

Solution- (i) Let y = (x+1/x).
We know: (x−1/x) = 8
Let’s square both sides:
(x−1/x)² = 8²
x²−2⋅x⋅1x+(1/x)² = 64
x²−2+1/x² = 64
x²+1/x² = 66
Now, let’s square (x+1/x) as well:
(x+1/x)² = y²
y² = x²+2+1/x²
We already know x+1/x² = 66:
y² = 66+2
y² = 68
y = √68​
y = ±√68
y = ±2√17
x+1/x = ​±2√17 Ans.

(ii) We already have:
x²+1/x² = 66
Using the identity for the difference of squares:
(x−1x)(x+1x) = x²−1/x²
Substituting the known values:
(8)(±2√17) = x²−1/x²
​x²−1/x² = ±16√17 Ans.

Que-24: If (x²+1/x²) = 7, find the values of (i) (x+1/x)  (ii) (x-1/x)
(iii) (2x² – 2/x²).

Solution- (i) Given
(x²+1/x² = 7)
x+1/x
Squaring we get
(x+1/x)² = x²+1/x²−2  (Using (a+b)²)
(x+1/x)² = (7+2)
(x+1/x)² = 9
Taking Square root on both sides , we get
(x+1x) = ±3 Ans.

(ii) Given
(x²+1/x² = 7)
x−1/x
Squaring we get
(x−1/x)² = x²+1/x²−2  (Using (a+b)²)
(x−1/x)² = (7−2)
(x−1/x)² = 5
Taking Square root on both sides , we get
(x−1x) = ±√5 Ans.

(iii) → x² + 1/x² = 7
Adding 2 both sides we get,
→ x² + 1/x² + 2 = 7 + 2
→ x² + 1/x² + 2 * x * 1/x = 9
Comparing LHS, with a²+b²+2ab = (a+b)² we get,
→ (x + 1/x)² = 9
Square root both sides we get,
→ (x + 1/x) = 3 ( Assuming Positive value only). ——- eq (1)
→ x² + 1/x² = 7
Subtracting 2 both sides we get,
→ x² + 1/x² – 2 = 7 – 2
→ x² + 1/x² – 2 * x * 1/x = 5
Comparing LHS, with a² + b² – 2ab = (a – b)² we get,
→ (x – 1/x)² = 5
Square root both sides we get,
→ (x – 1/x) = √5 ( Assuming Positive value only). eq (2)
→ 2(x² – 1/x²)
Using (a² – b²) = (a+b)(a – b) we get,
→ 2(x + 1/x)(x – 1/x)
Putting Values Form Equation (1) & (2) Now, we get,
→ 2 * 3 * √5
→ 6√5 (Ans).

Que-25: If (x²+1/25x²) = 9*(2/5), find the value of (x-1/5x)

Solution- (x²+1/25x²) = 9*(2/5) = 47/5
∵ (x+1/5x)² = (x)² + (1/5x)² − 2.x.1/5x
= x² + 1/25x² − 2/5
= 47/5 − 2/5
= (47−2)/5 = 45/5 = 9
⇒ x−1/5x = ±3 Ans.

Que-26: If (a²-4a-1) = 0 and a≠0, find the values of :
(i) (a-1/a)  (ii) (a+1/a)  (iii) (a²-1/a²)  (iv) (a²+1/a²)

Solution- (i) a² – 4a – 1 = 0
⇒ on taking a common we get,
⇒ a (a -4 -1/a ) = 0
⇒ a – 4 – 1/a = 0
⇒ a – 1/a = 4 Ans.

(ii) a² – 4a – 1 = 0
→ a² – 1 = 4a
→ a – (1/a) = 4 … (1)
Squaring on both sides
→ a² + (1/a)² – 2(a*1/a) = 16
→ a² + 1/a² = 16 + 2 = 18
Add 2(x*1/x) to both sides:
→ a² + 1/a² + 2(x*1/x) = 18 + 2(x*1/x)
→ (a + 1/a)² = 18 + 2 = 20
→ (a + 1/a) = ± √20 = ±√(4*5) = ±√(2² *5)
→ a + 1/a = ± 2√5 Ans.

(iii) a² – 4a – 1 = 0
→ a² – 1 = 4a
→ a – (1/a) = 4 … (1)
Squaring on both sides
→ a² + (1/a)² – 2(a*1/a) = 16
→ a² + 1/a² = 16 + 2 = 18
Add 2(x*1/x) to both sides:
→ a² + 1/a² + 2(x*1/x) = 18 + 2(x*1/x)
→ (a + 1/a)² = 18 + 2 = 20
→ (a + 1/a) = ± √20 = ±√(4*5) = ±√(2² *5)
→ a + 1/a = ± 2√5
a² – 1/a²
→ a² – (1/a)² = (a + 1/a)(a – 1/a)
→ (± 2√5)(4)
→ ± 8√5 Ans.

(iv) Now, on  squaring  both the sides we get,
( a – 1/a)² =   a² + 1/a² – 2 × a × 1/a
⇒ 16 = a² + 1/a² – 2
a² + 1/a² = 18 Ans.

Que-27: If a = 1/(a-5), where a≠5 and a≠0, find the values of :
(i) (a-1/a)  (ii) (a+1/a)  (iii) (a²-1/a²)  (iv) (a²+1/a²)

Solution- (i) a = 1/(a – 5)
⇒ a² – 5a = 1
Divide by a
⇒ a – 5 = 1/a
⇒ a – 1/a = 5 Ans.

(ii) a = 1/(a – 5)
⇒ a² – 5a = 1
Divide by a
⇒ a – 5 = 1/a
⇒ a – 1/a = 5
As we know
⇒ (a + 1/a)² – (a – 1/a)² = 4
⇒ (a + 1/a)² – 5² = 4
⇒ (a + 1/a)² = 25 + 4
∴ (a + 1/a) = √29 Ans.

(iii) (a²-1/a²)
using formula, (x² – y²) = (x – y)(x + y)
(a² – 1/a²) = (a – 1/a)(a + 1/a)
= 5√29
therefore, (a² – 1/a²) = 5√29 Ans.

(iv) (a²+1/a²)
using formula, x² + y² = (x + y)² – 2xy
so, (a² + 1/a²) = (a + 1/a)² – 2 × a × 1/a
= (a + 1/a)² – 2
= (√29)² – 2
= 29 – 2
= 27
therefore, (a² + 1/a²) = 27 Ans.

Que-28: Using (a+b)² = (a²+b²+2ab), evaluate :
(i) (137)²  (ii) (1008)²  (iii) (11.6)²

Solution- (i) (137)²
= (100+37)
This is in the form of (a+b)²
(a+b)² = a²+b²+2ab
a = 100 , b = 37
(100+37)² = (100)²+(37)²+2*100*37
= 10000+1369+200*37
= 11369+7400
= 18769 Ans.

(ii) (1008)²
(a+b) ² = a²+b² +2ab
(1008) ² = (1000+8) ²
= 1000² +8² + 2x 1000 x8
= 1000000 + 64 + 16000
= 1016064 Ans.

(iii) (11.6)²
(a+b)² = a²+b²+2ab
(11.6)² = (11+0.6)²
= (11)² + (0.6)² + 2 x 11 x 0.6
= 121 + 0.36 + 13.2
= 134.56 Ans.

Que-29: Using (a-b)² = (a²+b²-2ab), evaluate :
(i) (97)²  (ii) (992)²  (iii) (9.98)²

Solution- (97)²
(a-b)² = a²+b²-2ab
(97)² = (100-3)²
= (100)² + 3² – 2 x 100 x 3
= 10000 + 9 – 600
= 9409 Ans.

(ii) (992)²
(a-b)² = a²+b²-2ab
(992)² = (1000 – 8)²
= (1000)² + (8)² – 2 x 1000 x 8
= 1000000 + 64 – 16000
= 984064 Ans.

(iii) (9.98)²
(a-b)² = a²+b²-2ab
(9.98)² = (10-0.02)²
= (10)² + (0.02)² – 2 x 10 x 0.02
= 100 + 0.0004 – 0.4
= 99.6004 Ans.

Que-30: Fill in the blanks to make the given expression a perfect square :

(i) 16a² + 9b² +….  (ii) 25a² + 16b² – …   (iii) 4a² + 20ab + ….. (iv) 9a² – 24ab + ….

Solution- (i) 24ab
(ii) 40ab
(iii) 25b²
(iv) 16b²

Que-31: If (a+b+c) = 14 and (a²+b²+c²) = 74,find the value of (ab+bc+ca)

Solution- We know, (a + b + c)^² = a^² + b^² + c² + 2(ab + bc + ca)
=> (14)^² = 74 + 2(ab + bc + ca)
=> ab + bc + ca = 122 / 2 = 61 Ans.

Que-32: If (a+b+c) = 15 and (ab+bc+ca)= 74, find the value of (a²+b²+c²)

Solution-  We know, (a + b + c)² = a² + b² + c² = + 2(ab + bc + ca)
=> (15)^² = a² + b² + c² + 2 x 74
=> 225 = a² + b² + c² = + 148
= a² + b² + c² = 225 – 148 = 77
a² + b² + c² = 77 Ans.

Que-33: If (a²+b²+c²) = 50 and (ab+bc+ca)= 47, find the value of (a+b+c)

Solution- We know, (a + b + c)² = a² + b² + c² = + 2(ab + bc + ca)
=> (a + b + c)² = 50 + 2 x 47
=> (a + b + c)² = 50 + 94
= (a + b + c)² = 144
a + b + c = √144
a + b + c = ±12 Ans.

Que-34: If (a²+b²+c²) = 89 and (ab-bc-ca)= 16, find the value of (a+b-c)

Solution- Given : (a² + b² + c²) = 89
ab – bc – ca = 16
We know that;
⇒ (a + b + (-c))² = a² + b² + (-c)² + 2ab + 2(b)(-c) + 2(-c)(a)
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
⇒ (a + b + (-c))² = 89 + 2(ab – bc -ca)
⇒ (a + b – c)² = 89 + 2(16)
⇒ (a + b – c)² = 89 + 32
⇒ (a + b – c)² = 121
⇒ (a + b – c) = √121
⇒ (a + b – c) = 11 Ans.

– : End of Expansions Class 9 RS Aggarwal Exe-3A Goyal Brothers ICSE Foundation Maths Solutions : –

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

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