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NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities
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Exploring Algebraic Identities
Chapter – 4
| Exercise Set 4.1 |
1. Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:
(i) (7x + 4y)2
Ans: (7x + 4y)2 = (7x)2 + 2 × 7x × 4y + (4y)2
[∵ (a + b)2 = a2 + 2ab + b2] = 49x2 + 56xy + 16y2
(ii) (7/5x + 3/2y)2
Ans:
(iii) (2.5p + 1.5q)2
Ans: Using the identity
(a + b)2 = a2 + 2ab + b2
we expand (2.5p + 1.5q)2
= (2.5p)2 + 2(2.5p) (1.5q) + (1.5q)2
= 6.25p2 + 7.5pq + 2.25q2
(iv) (3/4s + 8t)2
Ans: (3/4s + 8t)2
= (3/4s)2 + 2(3/4s)(8t) + (8t)2
= 9/16s2 + 12st + 64t2
(v) (x + 1/2y)2
Ans: Using the identity
(a + b)2 = a2 + 2ab + b2
we expand
(x + 1/2y)2 = x2 + 2(x⋅1/2y) + (1/2y)2
= x2 + x/y + 1/4y2
(vi) (1/x + 1/y)2
Ans: (1/x + 1/y)2
= (1x)2 + 2(1x) (1y) + (1y)2
= 1/x2 + 2/xy + 1/y2
2. Using the same identity, find the values of the following:
(i) (64)2
Ans: Using the identity
(a + b)2 = a2 + 2ab + b2
Write 64 = 60 + 4
(60 + 4)2 = 602 + 2(60) (4) + 42
= 3600 + 480 + 16 = 4096
(ii) (105)2
Ans: (105)2
Using the identity: (a + b)2 = a2 + 2ab + b2
1052 = (100 + 5)2
= 1002 + 2 × 100 × 5 + 52
= 10000 + 1000 + 25
= 11025
(iii) (205)2
Ans: Using the identity
(a + b)2 = a2 + 2ab + b2
Write 205 = 200 + 5
(200 + 5)2 = 2002 + 2(200) (5) + 52
= 40000 + 2000 + 25 = 42025
| Exercise Set 4.2 |
1. Factor completely:
(i) 9x2 + 14xy + 16y2
Ans: 9x2 + 14xy + 16y2
= (3x)2 + 2 × (3x) × (4y) + (4y)2
= (3x + 4y)2 [∵ (a + b)2 = a2 + 2ab + b2]
(ii) 4s2 + 20st + 25t2
Ans: We use the identity
a2 + 2ab + b2 = (a + b)2
49x2 + 28xy + 4y2
= (7x)2 + 2(7x) (2y) + (2y)2 = (7x+2y)2
(iii) 49x2 + 28xy + 4y2
Ans: 49x2 + 28xy + 4y2
= (7x)2 + 2 × (7x) × (2y) + (2y)2
= (7x + 2y)2 [∵ (a + b)2 = a2 + 2ab + b2]
(iv) 64p2 + 32/3pq + 4/9q2
Ans: We use the identity
a2 + 2ab + b2 = (a + b)2
64p2 + 32/3pq + 49q2
= (8p)2 + 2(8p) (23q)↓ + (23q)2 = (8p + 23q)2
(v) 3a2 + 4ab + 43b2
Ans: 3a2 + 4ab + 43b2
= (√3a)2 + 2 × (√3a) × (2/√3b) + (2/√3b)2
[∵ (√3) = √3 × √3 = √3 × 3 = 3]
= (√3a + 2/√3b)2 [∵ (a + b)2 = a2 + 2ab + b2]
(vi) 9/5s2 + 6sv + 5v2
Ans: = (3/√5s)2 + 2 × (3/√5s) × (√5v) + (√5v)2
(∵ 6 = 2 × 3/√5 × √5)
= (3/√5s + √5v)2
[∵ (a + b)2 = a2 + 2ab + b2]
2. Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.
(i) (79)2
Ans: Using the identity
(a – b)2 = a2 – 2ab + b2
Write 79 = 80 – 1
(80 – 1)2 = 802 – 2(80) (1) + 12
= 6400 – 160 + 1 = 6241
(ii) (193)2
Ans: (193)2 = (200 – 1)2 = 2002 – 2(200) (7) + 72
= 40000 – 2800 + 49
= 37249
(iii) (299)2
Ans: Using the identity
(a – b)2 = a2 – 2ab + b2
Write 299 = 300 – 1
(300 – 1)2 = 3002 – 2(300) (1) + 12
= 90000 – 600 + 1 = 89401
| Exercise Set 4.3 |
1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 1172
Ans: 1172 = (100 + 10 + 7)2
= 1002 + 102 + 72 + 2(100)(10) + 2(10)(7) + 2(7)(100)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 10000 +100 + 49 + 2000 + 140 + 1400 = 13689
(ii) 782
Ans: We use identity:
(a – b)2 = a2 – 2ab + b2
782 = (80 – 2)2
= 802 – 2(80) (2) + 22 = 6400 − 320 + 4 = 6084
(iii) 1982
Ans: 1982 = (200 – 2)2
= 2002 + 22 – 2 × 200 × 2
[∵(a – b)2 = a2 – 2ab + b2]
= 40000 – 800 + 4
= 39204
(iv) 2142
Ans: We use identity:
(a + b)2 = a2 + 2ab + b2
2142 = (200 + 14)2
= 2002 + 2(200) (14) + 142
= 40000 + 5600 + 196 = 45796
(v) 11042
Ans: 11042 = (1000+ 100 + 4)2
= 10002 + 1002 + 42 + 2(1000) (100) + 2(100)(4) + 2(4)(1000)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816
(vi) 11202
Ans: We use identity:
(a + b)2 = a2 + 2ab + b2
11202 = (1100 + 20)2
= 11002 + 2(1100) (20) + 202 = 1210000 + 44000 + 400 = 1254400
2. Factor using suitable identities:
(i) 16y2 – 24y + 9
Ans: 16y2 – 24y + 9 = (4y)2 – 2 × (4y) × 3 + (3)2
= (4y – 3)2
[∵(a – b)2 = a2 – 2ab + b2]
(ii) 9/4s2 + 6st + 4t2
Ans: We use identity:
(a + b)2 = a2 + 2ab + b2
94s2 + 6st + 4t2
= (3/2s)2 + 2(3/2s) (2t) + (2t)2 = (3/2s + 2t)2
(iii) m2/9 + mk/3 + k2/4 + 3nk + 2mn + 9n2
Ans: = (m/3)2 + (k/2)2 + (3n)2 + 2 × (m/3) × (3n) + 2 × (k/2)×(m/3) + 2 × (k/2) × (3n)
= (m/3 + k/2 + 3n)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(iv) p2/16 – 2 + 16/p2
Ans: We use identity:
(a – b)2 = a2 – 2ab + b2
p2/16 – 2 + 16/p2
= (p/4)2 – 2(p/4 ⋅ 4/p) + (4p)2 = (p4 – 4p)2
(v) 9a2 + 4b2 + c2 – 12ab + 6ac – 46c
Ans: 9a2 + 4b2 + c2 – 12ab + 6ac – 46c
= (3a)2 + (-2b)2 + (c)2 + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)
= [3a + (-2b) + c]2
= (3a – 2b + c)2
3. Expand the following using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)2
Ans: Using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(p + 3q + 7r)2
Here a = p,b = 3q, c = 7r
= p2 + (3q)2 + (7r)2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
= p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr
(ii) (3x – 2y + 4z)2
Ans: (3x – 2y + 4z)2
= (3x)2 + (-2y)2 + (4z)2 + 2(3x)(-2y) + 2(-2y) + 2(-2y)(4z) + 2(3x)(4z)
= 9x2 + 4y2 + 16x2 – 12xy – 16yz + 24xz
4. Is this an identity?
(a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 2a2 + 2b2 + 2c2.
Ans: Let us verify by expanding the LHS.
(a + b – c)2 a2 + b2 + c2 + 2a2 + 2b2 + 2c2
(a – b + c)2 = a2 + b2 + c2 – 2ab + 2ac – 2bc
(a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
Adding all three:
LHS = 3a2 + 3b2 + 3c2 + (2ab – 2ab – 2ab)
+(-2ac + 2ac – 2ac) + (-2bc – 2bc + 2bc)
= 3a2 + 3b2 + 3c2 – 2ab – 2ac – 2bc
But RHS is:
2a2 + 2b2 + 2c2
Since
LHS ≠ RHS
the given expression is not an identity.
| Exercise Set 4.4 |
1. Fill in the blanks to complete the following identities:
(i) s2 – 11s + 24 = (__________) (__________)
Ans: We factor expression using the method
x2 + (a + b)x + ab = (x + a) (x + b).
s2 – 11s + 24
We need two numbers whose sum = -11 and product = 24: -3, -8
= (s – 3) (s – 8)
(ii) (__________) (x + 1) = (3x2 – 4x – 7)
Ans: (__________) (x + 1) = (3x2 – 4x – 7)
We have, 3x2 – 4x – 7
Multiply the coefficient of x2 (which is 3) by the constant term (which is -7): 3 x (-7) = -21
Now, we need to find two numbers that multiply to -21 and add up to the coefficient of x (which is -4).
The two numbers that satisfy this condition are -7 and 3 because (-7) x 3 = -21, (-7) + 3 = -4
Therefore, (3x2 – 4x -7) = 3x2 – (7 – 3)x – 7
= 3x2 + 3x – 7x – 7
= 3x(x + 1) – 7(x + 1)
= (3x – 7) (x + 1)
Thus, (3x – 7) (x + 1) = (3x2 – 4x – 7)
(iii) 10x2 – 11x – 6 = (2x – __) (__ + 2)
Ans: We factor expression using the method
x2 + (a + b)x + ab = (x + a) (x + b).
10x2 – 11x – 6 = (2x – ________) (________ + 2)
Try factors of 10x2 and -6:
= (2x – 3) (5x + 2)
So blanks = 3 and 5x
(iv) 6x2 + 7x + 2 = (__________) (__________)
Ans: 6x2 + 7x + 2 = (__________) (__________)
Multiply the coefficient of x2 (which is 6) by the constant term (which is 2): 6 × 2 = 12.
Now, we need to find two numbers that multiply to 12 and add up to the coefficient of x (which is 7). The two numbers that satisfy this condition are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.
Therefore, 6x2 + 7x + 2 = 6x2 + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2) (2x + 1).
Thus, 6x2 + 7x + 2 = (3x + 2) (2x + 1)
2. Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)2
Ans: We use identity (a + b)2
412 = (40 + 1)2
= 1600 + 80 + 1 = 1681
(ii) (27)2
Ans: (27)2 = (30 – 3)2 = 3022 × 30 × 3 + 32
= 900 – 180 + 9
= 729
[∵ (a – b)2 = a2 – 2ab + b2]
(iii) (23 × 17)
Ans: We use identity (a + b) (a – b)
23 × 17 = (20 + 3) (20 – 3)
= 202 – 32 = 400 – 9 = 391
(iv) (135)2
Ans: (135)2 = (140 – 5)2
= 1402 – 2 × 140 × 5 + 52
= 19600 – 1400 + 25
= 18225
[∵ (a – b)2 = a2 – 2ab + b2]
(v) (97)2
Ans: We use identity (a – b)2
972 = (100 – 3)2
= 10000 – 600 + 9 = 9409
(vi) (18 × 29)
Ans: (18 × 29) = (20 – 2) (20 + 9)
= 202 + (-2 + 9)20 + (-2) × 9
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= 400 + 140 – 18
= 522
(vii) (34 × 43)
Ans: We use identity (a – b)2
34 × 43 = ((34 + 43)/2)2 – ((43 – 34)/2)2
= 77/2, 9/2 ⇒ =(77/2)2 – (92)2
= 5929 – 81/4 = 5848/4 = 1462
(viii) (205)2
Ans: (205)2 = (205 + 5) (205 – 5) + 52
= 210 × 200 + 25
= 42025
[∵ a2 = (a + b)(a – b) + b2]
3. Factor the following:
(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
Ans: 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
= (3a)2 + (-b)2 + (2c)2 + 2(3a)(-b)
+2(3a)(2c) + 2(-b)(2c)
= (3a – b + 2c)2
(ii) 16s2 + 25t2 – 40st
Ans: 16s2 + 25t2 – 40st
= (4s)2 + (5t)2 – 2 × (4s)(5t)
= (4s – 5t)2
[∵ (a – b)2 = a2 + b2 – 2ab]
(iii) r2 – r – 42
Ans: r2 – r – 42
Find numbers: product = -42, sum = -1 → -7, 6
= (r – 7) (r + 6)
(iv) 49g2 + 14gb + h2
Ans: 49g2 + 14gb + h2
= (7g)2 + 2(7g)b + h2
= (7g + h)2 [∵ (a + b)2 = a2 + b2 + 2ab]
(v) 64u2 + 121v2 + 4W2 – 176uv – 32uw + 44vw
Ans: 64u2 + 121v2 + 4W2 – 176uv – 32uw + 44vw
= (8u)2 + (-11v)2 + (-2w)2 + 2(8u) (-11v)
+2(8u)(-2w) + 2(-11v)(-2w)
= (8u – 11v – 2w)2
| Exercise Set 4.5 |
1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
Ans:
We have,
3p2 – 3pq – 18q2 = 3 (p2 – pq – 6q2)
= 3(p2 – 3pq + 2pq – 6q2)
(∵ -6 = -3 × 2 and -3 + 2 = -1)
= 3[p(p – 3q) + 2q(p – 3q)]
3p2 – 3pq – 18q2 = 3(p + 2q)(p – 3q) …(i)
Now, p2 + 3pq – 10q2 = p2 + 5pq – 2pq – 10q2
[∵ -10 = 5 × (-2) and 5 + (-2) = 3]
= p(p + 5q) – 2q(p + 5q)
Therefore, p2 + 3pq – 10q2 = (p – 2q)(p + 5q) …(ii)
From (i) and (ii), we get
Ans:
Ans:
We have, w3 – v3 + x3 + 3wvx
= (w – v + x)[w2 + (-v)2 + x2 – xv(-v) – (-v)x – wx]
= (w – v + x)(w2 + v2 + x2 + vw + vx – wx)
[∵ x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – xz – yz)]
Now, w2 + v2 + x2 – 2wv – 2vx + 2wx = (w – v + x)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
Ans:
Ans:
We have,
x2 + x – 6- (x + 3)(x – 2)
x2 – 7x + 12 = (x – 3)(x – 4)
x2 – 6x + 8 = (x – 2)(x – 4)
x2 – 9 = (x – 3)(x + 3)
Therefore,
Ans:
| END-OF-CHAPTER EXERCISES |
1. Use suitable identities to find the following products:
(i) (-3x + 4)2
Ans: (-3x + 4)2
Using (a – b)2 = a2 – 2ab + b2
= (-3x)2 + 2(-3x)(4) + 42 = 9x2 – 24x + 16
(ii) (2s + 7) (2s – 7)
Ans: (2s + 7) (2s – 7)
We can use the identity
(a + b) (a – b) = a2 – b2
(25 + 7) (25 – 7) = (25)2 – (7)2
= 4s2 – 49
Ans: (p2 + 1/2) (p2 – 1/2)
= (p2)2 – (1/2)2 = p4 – 1/4
(iv) (2n + 7) (2n – 7)
Ans: (2n + 1) (2n – 1)
Using the identity: (a + b) (a – b) = a2 – b2
(2n + 7) (2n – 7) = (2n)2 – 72
= 4n2 – 49
(v) (s – 2t) (s2 + 2st + 4t2)
Ans: (s – 2t) (s2 + 2st + 4t2)
Using a3 – b3 = (a – b) (a2 + ab + b2)
= s3 – (2t)3 = s3 – 8t3
Ans: (1/2r – 4r)2
We simplify this by using the identity
(a – b)2 = a2 – 2ab + b2
(1/2r)2 – 2(1/2r)(4r) + (4r)2
= 1/4r2 – 4 + 16r2
(vii) (-3m + 4k – l)2
Ans: (-3m + 4k – l)2
Using (a + b + c)2
= 9m2 + 16k2 + l2 – 24mk + 6ml – 8kl
Ans: (x – 1/3y)3
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3
= x3 – x2y + 1/3xy2 – 1/27y3
Ans: To expand this, we use the identity:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
2. Find the values using suitable identities:
(i) 17 × 21
Ans: 17 × 21
Use (a – b) (a + b) = a2 – b2
= (19 – 2) (19 + 2) = 192 – 22 = 361 – 4 = 357
(ii) 104 × 96
Ans: 104 × 96
Here, 104 = 100 + 4 and 96 = 100 – 4,
So, 104 × 96 = (100 + 4) (100 – 4)
Using the identity (a + b) (a – b) = a2 – b2
= 1002 – 42
= 10000 – 16
= 9984
(iii) 24 × 16
Ans: 24 × 16
= (20 + 4) (20 – 4) = 202 – 42 = 400 – 16 = 384
(iv) 1473
Ans: 1473
We can use the identity for cubes:
(a – b)3 = a3 – 3a2b + 3 ab2 – b3
Here, 147 = 150 – 3,
So, 1473 = (150 – 3)3
= 1503 – 3 × 1502 × 3 + 3 × 150 × 32 – 33
Simplifying:
= 3375000 – 202500 + 4050 – 27
= 3176523
(v) 1993
Ans: 1993
= (200 – 1)3 = 2003 – 3(2002)(1) + 3(200)(1) – 1
= 8000000 – 120000 + 600 – 1 = 7880599
(vi) 1273
Ans: 1273
We can use the identity for cubes:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
Here, 127 = 130 – 3,
So, 1273 = (130 – 3)3
= 1303 – 3 × 1302 × 3 + 3 × 130 × 32 – 33
= 2197000 – 152100 + 3510 – 27
= 2048383
(vii) (-107)3
Ans: (-107)3 = -(1073)
1073 = (100 + 7)3 = 1000000 + 210000 + 14700 + 343 = 1225043
⇒ (-107)3 = -1225043
(viii) (-299)3
Ans: (-299)3
We can use the identity for cubes:
(a + b)3 = a3 + 3a2b + 3 ab2 + b3
Here, -299 = -300 + 1,
So, (-299)3 = (-300 + 1)3
= (-300)3 + 3 × (-300)2 × 1 + 3 × (-300) × 12+ 13
= -27000000 + 270000 – 900 + 1
= 26730899
3. Factor the following algebraic expressions:
(i) 4y2 + 1 + 1/16y2
Ans: 4y2 + 1 + 1/16y2
= (2y)2 + 2 × (2y) × (1/4y) + (1/4y)2
= (2y + 12y)2
[∵ (a + b)2 = a2 + 2ab + b2]
(ii) 9m2 – 1/25n2
Ans: 9m2 – 1/25n2
= (3m)2 – (1/5n)2
= (3m + 1/5n)(3m – 1/5n)
[∵ a2 – b2 = (a + b)(a – b)]
(iii) 27b3 – 1/64b3
Ans: 27b3 – 1/64b3
= (3b)3 – (1/4b)3 = (3b – 1/4b) (9b2 + 34 + 1/16b2)
(iv) x2 + 5x/6 + 1/6
Ans:
(v) 27u3 – 1/125 – 27u2/5 + 9u/25
Ans: 27u3 – 1/125 – 27u2/5 + 9u/25
= (3u)3 – (1/5)3 – 3(3u)2 (1/5) + 3(3u) (1/5)2
= (3u – 15)3
(vi) 64y3 – 1/125z3
Ans:
(vii) p3 + 27q3 + r3 – 9pqr
Ans: p3 + 27q3 + r3 – 9pqr
We have, x3 + y3 + z3 – 3xyz
= (x + y + z)(x3 + y3 + z3 – xy – xz – yz)
= (p + 3q + r)(p3 + 9q3 + r3 – 3pq – 3qr – pr)
(viii) 9m2 – 12m + 4
Ans: 9m2 – 12m + 4
= (3m)2 – 2(3m)(2) + 22 = (3m – 2)2
(ix) 9x3 – 8/3 y3 + z3/3 + 6xyz
Ans: 9x3 – 8/3 y3 + z3/3 + 6xyz
= 1/3(27x3 – 8y3 + z3 + 18xyz)
= 1/3[(3x)3 + (-2y)3 + (z)3 – 3(3x)(-2y)(z)]
= 1/3(3x – 2y +z)(9x3 + 4y3 + z3 + 6xy + 2yz – 3xz)
[∵ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
Ans: 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
= (2x)2 + (3y)2 + (6z)2 + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)2
(xi) 27u3 – 1/216 – 9u2/2 + u/4
Ans: 27u3 – 1/216 – 9u2/2 + u/4
= (3u)3 – (1/6)3 – 3(3u)2 (1/6) + 3(3u) (1/6)2
= (3u – 1/6)3
4. Simplify the following:
(i) 4x2 + 4x + 1/4x2 – 1
(ii) 9(3a3 – 24b3)/9a2 – 36b2
(iii) s3 + 125t3/s2 – 2st – 35t2
Note: Assume that the denominators are not equal to 0.
Ans:
5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a2 – 30ab + 9b2
Ans: Recognise identity: (a – b)2 = a2 – 2ab + b2
= (5a)2 – 2(5a) (3b) + (3b)2 = (5a – 3b)2
So,
Length = 5a – 3b
Breadth = 5a – 3b
(ii) 36s2 – 49t2
Ans: 36s2 – 49t2
= (6s)2 – (7t)2
= (6s + 7t)(6s – 7t)
[∵ a2 – b2 = (a + b) (a – b)]
Area of a rectangle = length × breadth
Possible dimensions:
Length = (6s + 7t) units
Breadth = (6s – 7t) units.
6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a2 – 24b2
Ans: 6a2 – 24b2
Take common factor:
= 6 (a2 – 4b2)
Use identity a2 – b2 = (a – b)(a + b):
= 6(a – 2b)(a + 2b)
So possible dimensions are:
Length = 6
Breadth = a – 2b
Height = a + 2b
(ii) 3ps2 – 15ps +12p
Ans: 3ps2 – 15ps +12p
= 3p(s2 – 5s + 4)
= 3p(s2 – 4s – s + 4)
= 3p[s(s – 4) – 1(s – 4)]
= 3p(s – 1)(s – 4)
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 3p units
Breadth = (s – 1) units
Height = (s – 4) units.
7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Ans: Side of square playground = 40 m
Path of width s is all around, so new outer square side:
= 40 + 2s
Area of outer square:
(40 + 2s)2
Area of playground:
402 = 1600
Area of path:
= (40 + 2s)2 – 1600
= (1600 + 160s + 4s2) – 1600
= 160s + 4s2
8. If a number plus its reciprocal equals 10/3, find the number.
Ans: Let the number be x. Then its reciprocal is 1x
According to question,
x + 1x = 103
⇒ x2 + 1x = 103
⇒ 3(x2 + 1) = 10x
⇒ 3x2 + 3 = 10x
⇒ 3x2 – 10x + 3 = 0
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1)(x – 3) = 0
⇒ x = 1/3 or x = 3
9. A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Ans: Area of rectangle = 2x2 + 7x + 3
Width = 2x + 1
We know:
Area = Length × Width
So,
Length = 2x2 + 7x + 3/2x + 1
Now factorise numerator:
2x2 + 7x + 3 = (2x + 1)(x + 3)
Cancel common factor:
Length = (2x + 1)(x + 3)/2x + 1 = x + 3
10. If both x – 2 and x – 1/2 are factors of px2 + 5x + r, show that p = r.
Ans: px2 + 5x + r
Factors = x – 2, x – 1/2
⇒ x = 2 or 1/2
For x = 2,
p(2)2 + 5(2) + r = 0
4p + 10 + r = 0 …(1)
For x = 1/2
p(1/2)2 + 5(1/2) + r = 0
p/4 + 5/2 + r = 0 …….(2)
Equating (1) and (2), we get
p = -2
Substituting p = -2 in equation (1), we get
4(-2) + 10 + r = 0
⇒ -8 + 10 + r = 0
⇒ 2 + r = 0
⇒ r = -2
∴ p = r = -2
Hence, p = r
11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25.
Ans: Given:
a + b + c = 5, ab + bc + ca = 10
Use the identity:
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
First find a2 + b2 + c2:
a2 + b2 + c2 = (a + b + c)2 – 2(ab + bc + ca)
= 52 – 2(10)
= 25 – 20 = 5
Now substitute:
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
= 5(5 – 10)
= 5 × (-5)
= -25
Final Result:
a3 + b3 + c3 – 3abc = -25
12. By factoring the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reasons.
Ans: n3 – n = n(n2 – 1)
Now using the identity a2 – b2 = (a + b)(a – b)
n3 – n = n(n – 1)(n + 1)
This is the product of three consecutive natural numbers.
We know that, among any three consecutive numbers:
One number is divisible by 3 (because every third number is a multiple of 3)
At least one number is even (divisible by 2)
So their product is divisible by 2 × 3 = 6
Therefore, n3 – n is always divisible by 6 for all natural numbers n.
Hence, proved.
13. Find the value of
(i) x3 + y3 – 12xy + 64, when x + y = -4
Ans: Given x + y = -4
Expression:
x3 + y3 – 12xy + 64
Use identity:
x3 + y3 = (x + y)3 – 3xy(x + y)
So,
x3 + y3 – 12xy + 64 = (x+y)3 – 3xy(x + y) – 12xy + 64
Substitute x + y = -4:
= (-4)3 – 3xy(-4) – 12xy + 64
=−64 + 12xy – 12xy + 64
= 0
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6
Ans: We are given, x = 2y + 6
⇒ x – 2y = 6
Now, x3 – 8y3 – 36xy – 216
Using the identity: (a – b)3 = a3 – b3 – 3ab(a – b)
(x – 2y)3 = x3 – (2y)3 – 3x(2y)(x – 2y)
(6)3 = x3 – 8y3 – 6xy(6)
216 = x3 – 8y3 – 36xy
⇒ x3 – 8y3 – 36xy – 216 = 0
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