ML Aggarwal Solutions Class 9 Math 3rd Chapter Expansions Exercise 3.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Third Chapter Expansions Exercise 3.2. APC Solution Class 9 Exercise 3.2.
(1) If x – y = 8 and xy = 5, find x2 + y2
Solution:
Given, x – y = 8, xy = 5
Or, (x – y)2 = 82
Or, x2 + y2 – 2xy = 64
Or, x2 + y2 = 64 + 2xy
Or, x2 + y2 = 64 + 2 × 5
Or, x2 + y2 = 74
(2) If x + y = 10 and xy = 21, find 2 (x2 + y2)
Solution:
Given, x + y = 10, xy = 21
Or, (x + y)2 = 102
Or, x2 + y2 + 2xy = 102
Or, x2 + y2 + 2× 21 = 100
Or, x2 + y2 = 100 – 42
Or, x2 + y2 = 58
Or, 2 (x2 + y2) = 58×2 [Multiplying 2 on both sides]
Or, 2 (x2 + y2) = 116
(3) If 2a + 3b = 7 and ab = 2, find 4a2 + 9b2
Solution:
Given, 2a + 3b = 7, ab = 2
Or, (2a + 3b)2= 72
Or, 4a2 + 9b2 + 2 × 2a × 3b = 49
Or, 4a2 + 9b2 + 12ab = 49
Or, 4a2 + 9b3 = 49 – 12 × 2
Or, 4a2 + 9b2 = 25
(4) If 3x – 4y = 16 and xy = 4, find the value of 9x2 + 16y2
Solution:
Given, 3x – 4y = 16, xy = 4
Or, (3x – 4y)2 = 162
Or, 9x2 + 16y2 – 2 × 3x × 4y = 256
Or, 9x2 + 16y2 – 24xy = 256
Or, 9x2 + 16y2 = 256 + 24 × 4
Or, 9x2 + 16y2 = 352
(5) If x + y = 8 and x – y = 2, find the value of 2x2 + 2y2
(Hint: 2 (x2 + y2) = (x + y)2 + (x – y)2
Solution:
x + y = 8, x – y = 2
2 (x2 + y2) = (x + y)2 + (x – y)2
Or, 2x2 + 2y2 = 82 + 22
Or, 2x2 + 2y2 = 16+4
Or, 2x2 + 2y2 = 20
(6) If a2 + b2 = 13 and ab = 6, find
(i) a + b
(ii) a – b
Solution:
(i) We know,
a2 + b2 = (a + b)2 – 2ab
Or, (a + b)2 = 13 + 2 × 6
Or, (a + b) = √25
Or, (a + b) = ±5
(ii) (a – b)2 = a2 + b2 – 2ab
Or, (a – b)2 = 13 – 2 × 6
Or a – b = √1
Or, a – b = ±1
(7) If a + b = 4 and ab = – 12, find
(i) a – b
(ii) a2 – b2
Solution:
a + b = 4, ab = -12
(i) (a – b)2 = a2 + b2 – 2ab
Adding 4ab on both sides we get.
(a – b)2 + 4ab = a2 + b2 – 2ab + 4ab
Or, (a – b)2 + 4ab = a2 + b2 + 2ab
Or, (a – b)2 + 4 × -12 (a + b)2
Or, (a – b)2 – 48 = 42
Or, (a – b)2 = 16 + 48
Or, (a – b) = √64
Or, a – b = ±8
(ii) a2 – b2 = (a + b) (a – b)
= 4 ×±8
= ±32
(8) If p – q = 9 and pq = 36, Evaluate
(i) p + q
(ii) p2 – q2
Solution:
p – q = 9, pq = 36
(i) p – q = q
Or, (p – q)2 = q2
Or, p2 + q2 – 2pq = 81
Or, p2 + q2 – 2pq + 4pq = 81 + 4pq [Adding equation on both sides]
Or, p2 + q2 + 2pq = 81 + 4pq
Or, (p + q)2 = 81 + 4 × 36
Or, p + q = √225
Or, p + q = ±15
(ii) We know, (p + q) (p – q) = p2 – q2
Or, p2 – q2 = ±15 × 9
Or, p2 – q2 = ± 135
(9) If x + y = 6 and x – y = 4, find
(i) x2+ y2
(ii) xy
Hint: (ii) 4xy = (x + y)2 – (x – y)2
Solution:
x + y = 6, x – y = 4
Or, (x2 + y)2 = 62
Or, x2 + y2 + 2xy = 36
Or, x2 + y2 = 36 – 2xy —- (i)
Also, x – y = 4
Or, (x – y)2 = 42
Or, x2 + y2 – 2xy = 16
Or, x2 + y2 = 16 + 2xy —– (ii)
Comparing equation (i) & (ii) we get,
36 – 2xy = 16 + 2xy
Or, 2xy + 2xy = 36 – 16
Or, xy = 20/4
Or, xy = 5
(i) Putting the value of xy in equation (i) we get,
x2 + y2 = 36 – 2xy
Or, x2 + y2 = 36 – 2xy
Or, x2 + y2 = 36 – 2 × 5
Or, x2 + y2 = 36 – 10
Or, x2 + y2 = 26
(10) If x – 3 = 1/x, find the value of x2 + 1/x2
Hint: x – 3 = 1/x => x – 1/x = 3.
Solution:
x – 3 = 1/x
Or, x – 1/x = 3
Or, (x – 1/x)2 = 32
Or, x2 + 1/x2 – 2 × X × 1/x = 9
Or, x2 + 1/x2 = 9+2
Or, x2 + 1/x2 = 11
(11) If x + y = 8 and xy = 3 3/4, find the values of
(i) x – y
(ii) 3 (x2 + y2)
(iii) 5 (x2 + y2) + 4 (x – y)
Solution:
x + y = 8,
xy = 3 3/4 = 15/4
(i) (x + y)2 = 82
Or, x2 + y2 + 2xy = 64
Or, x2 + y2 + 2xy – 4xy = 64 – 4xy
Or, x2 + y2 – 2xy = 64 – 4 × 15/4
Or, (x – y)2 = 49
Or, x – y = √49
Or, x – y = ± 7
(ii) x + y = 8
Or, (x + y)2 = 82
Or, x2 + y2 + 2xy = 64
Or, x2 + y2 = 64 – 2 × 15/4
Or, x2 + y2 = 128-15/2
Or, x2 + y2 = 113/2
Or, 3 (x2 + y2) = 113/2 × 3
Or, 3 (x2 + y2) = 339/2
(iii) 5 (x2 + y2) + 4 (x – y)
= 5 × 113/2 + 4 ×±7
= 565/2 ± 28
either, = 565+456/2
= 621/2
Or, = 565/2 – 28
= 565-56/2
= 509/2
(12) If x2 + y2 = 34 and xy = 10 1/2, find the value of 2 (x + y)2 + (x – y)2
Solution:
x2 + y2 = 34, xy = 10 1/2 = 21/2
Or, x2 + y2 + 2xy = 34 + 2xy
Or, (x + y)2 = 34 + 2 × 21/2
Or, x + y = √55
Also, x2 + y2 = 34
Or, x2 + y2 – 2xy = 34 – 2xy
Or, (x – y)2 = 34 – 2 × x/2
Or, x – y = √13
∴ 2 (x + y)2 + (x – y)2
= 2 (√55)2 + (√13)2
= 2 × 55 + 13
= 110 + 13
= 123
(13) If a – b = 3 and ab = 4, find a3 – b3
Solution:
a – b = 3, ab = 4
Or, (a – b)3 =33
Or, a3 – b3 – 3ab (a – b) = 27
Or, a3 – b3 – 3×4×3 = 27
Or, a3 – b3 = 27+36
Or, a3 – b3 = 63
(14) If 2a – 3b = 3 and ab = 2, find the value of 8a3 – 27b3.
Solution:
2a – 3b = 3, ab = 2
Or, (2a – 3b)3 = 33
Or, 8a3 – 27b3 – 3 × 2a.3b (2a – 3b) = 27
Or, 8a3 – 27b3 – 18ab (2a – 3b) = 27
Or, 8a2 – 27b3 = 27 + 18 × 2 × 3
Or, 8a3 – 27b3 = 27 + 108
Or, 8a3 – 27b3 = 135
(15) If x + 1/x = 4, find the values of:
(i) x2 + 1/x2
(ii) x4 + 1/x4
(iii) x3 + 1/x3
(iv) x – 1/x
Solution:
x + 1/x = 4
(i) x + 1/x = 4
Or, (x + 1/x)2 = 42
Or, x2 + 1/x2 + 2 = 16
Or, x2 + 1/x2 = 14
(ii) x + 1/x = 4
= (x + 1/x)2 = 42
Or, x2 + 1/x2 + 2 = 16
Or, x2 + 1/x2 = 14
Or, (x2 + 1/x2)2 = 142
Or, x4 + 1/x2 + 2 = 196
Or, x2 + 1/x2 = 194
(iii) a + 1/x = 4
Or, (x + 1/x)3 = 43
Or, x3 + 1/x3 + 3.x.1/x (x + 1/x) = 64
Or, x3 1/x3 + 3 × 4 = 64
Or, x3 + 1/x3 = 64 – 12
Or, x3 + 1/x3 = 52
(iv) x + 1/x = 4
Or, (x + 1/x)2 = 42
Or, x2 + 1/x2 + 2 = 16
Or, x2 + 1/x2 + 2 – 4 = 16 – 4
Or, x2 + 1/x2 – 2 = 12
Or, (x – 1/x)2 = 12
Or, x – 1/x = √12
Or, x – 1/x = ±2√3
(16) If x – 1/x = 5, find the value of x4 + 1/x4
Solution:
x – 1/x = 5
Or, (x – 1/x)n = 52
Or, x2 + 1/x2 – 2 = 25
Or, (x3 + 1/x2)n = 272
Or, x4 + 1/x2 + 2 = 729
Or, x4 + 1/x4 = 729 – 2
Or, x4 + 1/x4 = 727
(17) If x – 1/x = √5, fin the values of
(i) x2 + 1/x2
(ii) x + 1/x
(iii) x3 + 1/x3
Solution:
x – 1/x = √5
(i) (x – 1/x)2 (√5)2
Or, x2 + 1/x2 – 2 = 5
Or, x2 + 1/x2 = 7
(ii) x2 + 1/x2 = 7
Or, x2 + 1/x2 + 2 = 7+2
Or, (x + 1/x)3 = 9
Or, x + 1/x = √9
Or, x + 1/x = ±3
(iii) (x + 1/x)3= (±3)3
Or, x3 + 1/x3 + 3 (x + 1/x) = ±27
Or, x3 + 1/x3 + 3 × ± 3 = ± 27
Or, x3 + 1/x3 = ± 27 – 9
Or, x3 + 1/x3 = ± 18
(18) If x + 1/x = 6, find
(i) x – 1/x
(ii) x2 – 1/x2
Solution:
x + 1/x = 6
(i) x + 1/x = 6
Or, (x + 1/x)2 = 62
Or, x2 + 1/x2 + 2 = 36
Or, x2 + 1/x2 – 2 = 36 – 4
Or, (x – 1/x)2 = 32
Or, x – 1/x = √32 = ±4√2
(ii) (x + 1/x) (x – 1/x) = x2 – 1/x2
Or, x3 – 1/x2 = 6 ×± 4√2
Or, x2 – 1/x2 = ± 24√2
(19) If x + 1/x = 2, Prove that x2 + 1/x2 = x3 + 1/x3 = x+ 1/x4
Solution:
x + 1/x = 2
Or, (x + 1/x)n = 22
Or, x2 + 1/x2 + 2 = 4
Or, x2 + 1/x2 = 2 —- (i)
x2 + 1/x2 = 2
Or, (x2 + 1/x2)2 = 22
Or, x4 + 1/x4 + 2 = 4
Or, x4 + 1/x4 = 2 —- (ii)
(x + 1/x)3 = 23
Or, x3 + 1/x3 + 3 (x + 1/x) = 8
Or, x3 + 1/x3 + 3 × 2 = 8
Or, x3 + 1/x3 = 8 – 6
Or, x3 + 1/x3 = 2 —– (iii)
Comparing equation (i), (ii), & (iii) we get,
x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4 = 2 (Proved)
(20) If x – 2/x = 3, find the value of x3 – 8/x3
Solution:
x – 2/x = 3
Or, (x – 2/x)3 = 33
Or, x3 – 8/x3 – 3 × x.2/x (x – 2/x) = 27
Or, x3 – 8/x3 – 6 × 3 = 27
Or, x3 – 8/x3 = 27 + 18
Or, x3 = 8/x3 = 45
(21) If a + 2b = 5, Prove that a3 + 8b3 + 30ab = 125
Solution:
a + 2b = 5
Or, (a + 2b)3 = 53
Or, a3 + 8b3 + 3 × a.2b (a + 2b) = 125
Or, a3 + 8b3 + 36ab × 5 = 125
Or, a3 + 8b3 + 30ab = 125 Proved
(22) If a + 1/a = p, Prove that a3 + 1/a3 = p (p2 – 3).
Solution:
a + 1/a = p
Or, (a + 1/a)3 = p3
Or, a3 + 1/a3 + 3 (a + 1/a) = p3
Or, a3 + 1/a3 = p3 – 3 × p
Or, a3 + 1/a3 = p (p3 – 3) (Proved)
(23) If x2 + 1/x2 = 27, find the value of x – 1/x.
Solution:
x2 + 1/x2 = 27
Or, x2 + 1/x2 – 2 = 27 – 2
Or, (x – 1/x)2 = 25
Or, x – 1/x = √25
Or, x – 1/x = ±5
(24) If x2 + 1/x2 = 27, find the value of 3x3 + 5x – 3/x3 – 5/x
Solution:
X2 + 1/x2 = 27
Or, x2 + 1/x2 – 2 = 27 – 2
Or, (x – 1/x)2 = 25
Or, x – 1/x = √25
Or, x – 1/x = ± 5 —– (i)
Or, (x – 1/x)3 = 53
Or, x3 – 1/x3 – 3 (x – 1/x) = 125
Or, x3 – 1/x3 = 125 + 3 × 5
Or, x3 – 1/x3 = 140 —- (ii)
Now, 3x3 + 5x – 3/x3 – 5/x
= 3x3 – 3/x+ 5x – 5/x
= 3 (x3 – 1/x3) + 5 (x – 1/x)
= 3 × 140 + 5 × 5 [From (i) & (ii)]
= 420 + 25
= ± 445
(25) If x2 + 1/25x2 = 8 3/5, find x + 1/5x
Solution:
x2+ 1/25x2 = 8 3/5
Or, x2 + 1/25x2 + 2/5 = 43/5 + 2/5
Or, x2 + 1/25x2 + 2 × x × 1/5x = 43+2/5
Or, (x + 1/5x)n = 45/5
Or, x + 1/5x = √9
Or, x + 1/5x = ± 3
(26) If x2 + 1/4x2 = 8, find x3 + 1/8x3
Solution:
x2 + 1/4x2 = 8
Or, x2 + 1/4x2 + 1 = 8 + 1
Or, x2 + 1/4x2 + 2.x.1/2x = 9
Or, (x + 1/2x)2 = 9
Or, x + 1/2x = √9
= ±3
∴ (x + 1/2x)3 = 33
Or, x3 + 1/8x3 + 3/2 (x + 1/2x) = 27
Or, x3 + 1/8x3 + 3/2 × 3 = 27
Or, x3 + 1/8x3 = 27 – 9/2
Or, x3 + 1/8x3 = 45/2
Or, x3 + 1/8x3 = ± 22 1/2
(27) If a2 – 3a + 1 = 0, find
(i) a2 + 1/a2
(ii) a3 + 1/a3
Solution:
a2 – 3a + 1 = 0
Or, a (a – 3 + 1/a) = 0
Or, a – 3 + 1/a = 0
Or, a + 1/a = 3
(i) a + 1/a = 3
Or, (a + 1/a)2 = 32
Or, a2 + 1/a2 + 2 = 9
Or, a2 + 1/a2 = 7
(ii) (a + 1/a)3 = 33
Or, a3 + 1/a3 + 3 (a + 1/a) = 27
Or, a3 + 1/a3 + 3 × 3 = 27
Or, a3 + 1/a3 = 27 – 9
Or, a3 + 1/a3 = 18
(28) If a = 1/a-5, find
(i) a – 1/a
(ii) a + 1/a
(iii) a2 – 1/a2
Hint: a = 1/a-5 => a2 – 5a – 1 = 0
Solution:
a = 1/a-5
Or, a (a – 5) = 1
Or, a2 – 5a – 1 = 0
Or, a (a – 5 – 1/a) = 0
Or, a – 1/a = 5
(i) a – 1/a = 5
(ii) (a – 1/a)2 = 52
Or, a2 + 1/a2 – 2 = 25
Or, a2 + 1/a2 + 2 = 25+4
Or, (a + 1/a)2 = 29
Or, a2 + 1/a = ±√29
(iii) a2 – 1/a2 = (a + 1/a) (a – 1/a)
Or, a2 – 1/a2 = ±√4 × 5
Or, a2 – 1/a2 = ± 5√29
(29) If (x + 1/x)2 = 3, find x3 + 1/x3.
Solution:
(x + 1/x)2 = 3
Or, (x + 1/x) = √3 — (i)
Or, (x + 1/x)3 = (√3)3
Or, x3 + 1/x3 + 3 (x + 1/x) = 3√3
Or, x3 + 1/x3 + 3√3 = 3√3 [From (i)]
Or, x3 + 1/x3 = 3√3 – 3√3
Or, x3 + 1/x3 = 0
30)
(31) If a + b + c = 12 and ab + bc + ca = 22, find a2 + b2 + c2.
Solution:
a + b + c = 12, ab + bc + ac = 22.
Or, (a + b + c)2 = 122
Or, a2 + b2 + c2 + 2 (ab + bc + ac) = 144
Or, a2 + b2 + c2 + 2 × 22 = 144
Or, a2 + b2 + c2 = 144 – 44
Or, a2 + b2 + c2 = 100
(32) If a + b + c = 12 and a2 + b2 + c2 = 100, find ab + bc + ca
Solution:
a + b + c = 12, a2 + b2 + c2 = 100
Or, (a + b + c)2 = 122
Or, a2 + b2 + c2 + 2 (ab + bc + ac) = 144
Or, 100 + 2 (ab + bc + ac) = 144
Or, 2 (ab + bc + ac) = 144 – 100
Or, ab + bc + ac = 44/2
Or, ab + bc + ac = 22
(33) If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.
Solution:
a2 + b2 + c2 = 125, ab + bc + ac = 50
We know, (a + b + c)2 = a2 + b2 + c2 + (ab + bc + ca)
Or, (a + b + c)2 = 125 + 2 × 50
Or, (a + b + c)n = 225
Or, a + b + c = √225
Or, a + b + c = 15
(34) If a + b – c = 5 and a2 + b2 + c2 = 29, find the value of ab – bc – ca
Solution:
a + b – c = 5, a2 + b2 + c2 = 29
Or, (a + b – c)2 = 52
Or, (a + b – c) (a + b – c) = 25
Or, a2 + ab – ac + ab + b2 – bc – ac – bc + c2 = 25
Or, a2 + b2 + c2 + 2ab – 2bc – 2ac = 25
Or, 29 + 2 (ab – bc – ac) = 25
Or, ab – bc – ac = 25-29/2
Or, ab – bc – ac = – 4/2
Or, ab – bc – ac = -2
(35) If a – b = 7 and a2 + b2 = 85, then find the value of a3 – b3.
Solution:
a – b = 7, a2 + b2 = 85
Or, (a – b)2 = 72
Or, a2 + b2 – 2ab = 49
Or, 85 – 26 = 49
Or, – 2ab = 49 – 85
Or, – 2ab = -36
Or, ab = 18 —- (i)
∴ (a – b)3 = 73
Or, a3 – b3 – 3ab (a – b) = 343
Or, a3 – b3– 3 × 18 × 7 = 343 [From (i)]
Or, a3 – b3 = 343 + 378
Or, a3 – b3 = 721
(36) If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.
Solution:
y – x = 3, y2 + x2 = 29
Or, (y – x)3 = 33
Or, y2 + x2 – 2yx = 9
Or, 29 – 2yx = 9
Or, – 2yx = 9 – 20
Or, yx = 20/2
Or yx = 10
∴ Product of x & y is 10.
(37) If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.
Solution:
Let, the two no. be x & y
x + y = 8, xy = 15
Or, (x + y)3 = 83
Or, x3 + y3 + 3xy (x + y) = 512
Or, x3 + y3 + 3 × 15 × 8 = 512
Or, x3 + y3 = 512 – 360
Or, x3 + y3 = 152
∴ Sum of their cubes is 152.
Multiple choice questions:
(1) If x + 1/x = 2, then x2 + 1/x2 is equal to
(a) 4
(b) 2
(c) 0
(d) None of these
Solution:
x + 1/x = 2
Or, (x + 1/x)2 = 22
Or, x2 + 1/x2 + 2 = 4
Or, x2 + 1/x2 = 2
Ans: Option (b) 2
(2) If x2 + y2 = 9 and xy = 8, then x + y is equal to
(a) 25
(b) 5
(c) -5
(d) ± 5
Solution:
x2 + y2 = 9, xy = 8
Or, x2 + y2 + 2xy = 9 + 2xy
Or, (x2 + y)2 = 9 + 2 × 8
Or, x + y = √25 = ± 5
Ans: Option (d) ± 5
(3) (102)2 – (98)2 is equal to
(a) 200
(b) 400
(c) 600
(d) 800
Solution:
(102)2 – (98)2 = (102 + 98) (102 – 98) = 200 × 4 = 800
Ans – Option (d) 800.
(4) 96 × 104 is equal to
(a) 9984
(b) 9974
(c) 9964
(d) None of these
Solution:
96 × 104 = (100 – 4) (100 + 4) = 1002 – 42 = 10000 – 16
= 9984
Ans – Option (a) 9984
(5) 1032 – 972/200 is equal to
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
1032-972/200 = (103+47)(103-97)/200 = 200×6/200 = 6
Ans: Option (d) 6.
(6) If x + y = 11 and xy = 24, then x2 + y2 is equal to
(a) 121
(b) 73
(c) 48
(d) 169
Solution:
x + y = 11, xy = 24 (x+y)2 = 112 => x2 + y2 + 2 × 24 = 121
x2 + y2 = 73
Option (b)
(7) The value of 2492 – 2482 is
(a) 12
(b) 477
(c) 487
(d) 497
Solution:
2492 – 2482 = (249+248) (249-248) = 497×1 = 497
Option – (d)