Chapter 4: Exploring Algebraic Identities
Complete Exercise Solutions Manual | Grade 9 Mathematics (Ganita Manjari)
Key Identities Covered:
- (a + b)2 = a2 + 2ab + b2
- (a - b)2 = a2 - 2ab + b2
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
- a2 - b2 = (a + b)(a - b)
- (x + a)(x + b) = x2 + (a + b)x + ab
- (a + b)3 = a3 + 3a2b + 3ab2 + b3
- (a - b)3 = a3 - 3a2b + 3ab2 - b3
- a3 + b3 = (a + b)(a2 - ab + b2)
- a3 - b3 = (a - b)(a2 + ab + b2)
- a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Exercise Set 4.1
1. Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:
(i) (7x + 4y)2
Using identity with a = 7x, b = 4y:
= (7x)2 + 2(7x)(4y) + (4y)2
Answer: 49x2 + 56xy + 16y2
(ii) (🍇x + 🍇y)2 → ( 7/5 x + 3/2 y )2
Using identity with a = 7/5 x, b = 3/2 y:
= ( 7/5 x )2 + 2( 7/5 x )( 3/2 y ) + ( 3/2 y )2
Answer: 49/25 x2 + 21/5 xy + 9/4 y2
(iii) (2.5p + 1.5q)2
Using identity with a = 2.5p, b = 1.5q:
= (2.5p)2 + 2(2.5p)(1.5q) + (1.5q)2
Answer: 6.25p2 + 7.5pq + 2.25q2
(iv) ( 3/4 s + 8/2 t )2 = ( 3/4 s + 4t )2
Using identity with a = 3/4 s, b = 4t:
= ( 3/4 s )2 + 2( 3/4 s )( 4t ) + ( 4t )2
Answer: 9/16 s2 + 6st + 16t2
(v) ( x/y + 1/2 )2
= ( x/y )2 + 2( x/y )( 1/2 ) + ( 1/2 )2
Answer: x2/y2 + x/y + 1/4
(vi) ( 11 + 2/xy )2
= 112 + 2(11)( 2/xy ) + ( 2/xy )2
Answer: 121 + 44/xy + 4/x2y2
2. Using the same identity, find the values of the following:
(i) (64)2 = (60 + 4)2
= 602 + 2(60)(4) + 42 = 3600 + 480 + 16
Answer: 4096
(ii) (105)2 = (100 + 5)2
= 1002 + 2(100)(5) + 52 = 10000 + 1000 + 25
Answer: 11025
(iii) (205)2 = (200 + 5)2
= 2002 + 2(200)(5) + 52 = 40000 + 2000 + 25
Answer: 42025
Exercise Set 4.2
1. Factor completely:
(i) 9x2 + 24xy + 16y2
= (3x)2 + 2(3x)(4y) + (4y)2
Answer: (3x + 4y)2
(ii) 4s2 + 20st + 25t2
= (2s)2 + 2(2s)(5t) + (5t)2
Answer: (2s + 5t)2
(iii) 49x2 + 28xy + 4y2
= (7x)2 + 2(7x)(2y) + (2y)2
Answer: (7x + 2y)2
(iv) 64p2 + 32/3 pq + 4/9 q2
= (8p)2 + 2(8p)( 2/3 q ) + ( 2/3 q )2
Answer: (8p + 2/3 q)2
(v) 3a2 + 4ab + 4/3 b2
Factor out 1/3: = 1/3 (9a2 + 12ab + 4b2)
= 1/3 [(3a)2 + 2(3a)(2b) + (2b)2]
Answer: 1/3 (3a + 2b)2
(vi) 9/5 s2 + 6sv + 5v2
Factor out 1/5: = 1/5 (9s2 + 30sv + 25v2)
= 1/5 [(3s)2 + 2(3s)(5v) + (5v)2]
Answer: 1/5 (3s + 5v)2
2. Find the values of the following using the identity (a - b)2 = a2 - 2ab + b2:
(i) (79)2 = (80 - 1)2
= 802 - 2(80)(1) + 12 = 6400 - 160 + 1
Answer: 6241
(ii) (193)2 = (200 - 7)2
= 2002 - 2(200)(7) + 72 = 40000 - 2800 + 49
Answer: 37249
(iii) (299)2 = (300 - 1)2
= 3002 - 2(300)(1) + 12 = 90000 - 600 + 1
Answer: 89401
Exercise Set 4.3
1. Find the following squares using suitable identities:
(i) 1172 = (100 + 17)2 or (100 + 10 + 7)2
Using (a+b)2 = (120 - 3)2: 1202 - 2(120)(3) + 32 = 14400 - 720 + 9
Answer: 13689
(ii) 782 = (80 - 2)2
= 802 - 2(80)(2) + 22 = 6400 - 320 + 4
Answer: 6084
(iii) 1982 = (200 - 2)2
= 2002 - 2(200)(2) + 22 = 40000 - 800 + 4
Answer: 39204
(iv) 2142 = (200 + 14)2
= 2002 + 2(200)(14) + 142 = 40000 + 5600 + 196
Answer: 45796
(v) 11042 = (1000 + 104)2 = (1000 + 100 + 4)2
Using (1000 + 104)2 = 1000000 + 208000 + 10816
Answer: 1218816
(vi) 11202 = (1000 + 120)2
= 10002 + 2(1000)(120) + 1202 = 1000000 + 240000 + 14400
Answer: 1254400
2. Factor using suitable identities:
(i) 16y2 - 24y + 9
= (4y)2 - 2(4y)(3) + 32
Answer: (4y - 3)2
(ii) 9/4 s2 + 6st + 4t2
= ( 3/2 s )2 + 2( 3/2 s )( 2t ) + (2t)2
Answer: ( 3/2 s + 2t )2
(iii) m2/9 + mk/3 + k2/4 + 3nk + 2mn + 9n2
Rearranging terms: ( m/3 )2 + ( k/2 )2 + (3n)2 + 2( m/3 )( k/2 ) + 2( k/2 )(3n) + 2( m/3 )(3n)
Answer: ( m/3 + k/2 + 3n )2
(iv) p2/16 - 2 + 16/p2
= ( p/4 )2 - 2( p/4 )( 4/p ) + ( 4/p )2
Answer: ( p/4 - 4/p )2
(v) 9a2 + 4b2 + c2 - 12ab + 6ac - 4bc
= (-3a)2 + (2b)2 + (-c)2 + 2(-3a)(2b) + 2(2b)(-c) + 2(-3a)(-c) (or equivalent sign variants)
Answer: (3a - 2b - c)2 or (-3a + 2b + c)2
3. Expand using (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca:
(i) (p + 3q + 7r)2
= p2 + (3q)2 + (7r)2 + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)
Answer: p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr
(ii) (3x - 2y + 4z)2
= (3x)2 + (-2y)2 + (4z)2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(3x)(4z)
Answer: 9x2 + 4y2 + 16z2 - 12xy - 16yz + 24xz
4. Is this an identity? (a+b-c)2 + (a-b+c)2 + (a-b-c)2 = 2a2 + 2b2 + 2c2?
Expanding LHS:
1) (a+b-c)2 = a2 + b2 + c2 + 2ab - 2bc - 2ca
2) (a-b+c)2 = a2 + b2 + c2 - 2ab - 2bc + 2ca
3) (a-b-c)2 = a2 + b2 + c2 - 2ab + 2bc + 2ca
Summing all three:
LHS = 3a2 + 3b2 + 3c2 - 2ab - 2bc + 2ca
Since 3a2 + 3b2 + 3c2 - 2ab - 2bc + 2ca ≠ 2a2 + 2b2 + 2c2, it is NOT an identity.
Answer: No, it is not an identity.
Exercise Set 4.4
1. Fill in the blanks:
(i) s2 - 11s + 24 = (s - 3)(s - 8)
(ii) (3x - 7)(x + 1) = 3x2 - 4x - 7
(iii) 10x2 - 11x - 6 = (2x - 3)(5x + 2)
(iv) 6x2 + 7x + 2 = (2x + 1)(3x + 2)
2. Find products without multiplying directly:
(i) (41)2 = (40 + 1)2 = 1600 + 80 + 1 = 1681
(ii) (27)2 = (30 - 3)2 = 900 - 180 + 9 = 729
(iii) (23 × 17) = (20 + 3)(20 - 3) = 202 - 32 = 400 - 9 = 391
(iv) (135)2 = (130 + 5)2 = 16900 + 1300 + 25 = 18225
(v) (97)2 = (100 - 3)2 = 10000 - 600 + 9 = 9409
(vi) (18 × 29) = (20 - 2)(30 - 1) = 600 - 20 - 60 + 2 = 522
(vii) (34 × 43) = (40 - 6)(40 + 3) = 1600 + 120 - 240 - 18 = 1462
(viii) (205)2 = (200 + 5)2 = 40000 + 2000 + 25 = 42025
3. Factor the following:
(i) 9a2 + b2 + 4c2 - 6ab + 12ac - 4bc
Answer: (-3a + b + 2c)2 or (3a - b - 2c)2
(ii) 16s2 + 25t2 - 40st = (4s)2 - 2(4s)(5t) + (5t)2
Answer: (4s - 5t)2
(iii) r2 - r - 42
Find numbers that multiply to -42 and add to -1: -7 and 6.
Answer: (r - 7)(r + 6)
(iv) 49g2 + 14gh + h2 = (7g)2 + 2(7g)(h) + h2
Answer: (7g + h)2
(v) 64u2 + 121v2 + 4w2 - 176uv - 32uw + 44vw
= (-8u)2 + (11v)2 + (2w)2 + 2(-8u)(11v) + 2(11v)(2w) + 2(-8u)(2w)
Answer: (-8u + 11v + 2w)2 or (8u - 11v - 2w)2
Exercise Set 4.5
1. Simplify the rational expressions:
(i) (3p2 - 3pq - 18q2) / (p2 + 3pq - 10q2)
Numerator: 3(p2 - pq - 6q2) = 3(p - 3q)(p + 2q)
Denominator: (p + 5q)(p - 2q) (Note: If denominator is p2 - 3pq - 10q2 = (p - 5q)(p + 2q), cancellation occurs). Assuming standard book text: p2 - 3pq - 10q2:
If denominator is (p - 5q)(p + 2q): 3(p - 3q)(p + 2q) / (p - 5q)(p + 2q) = 3(p - 3q) / (p - 5q)
Answer: 3(p - 3q) / (p - 2q) or 3(p - 3q) / (p - 5q) depending on sign in original text.
(ii) (n3 - 3n2m + 3nm2 - m3) / (5m2 - 10mn + 5n2)
Numerator: (n - m)3
Denominator: 5(m2 - 2mn + n2) = 5(n - m)2
= (n - m)3 / 5(n - m)2
Answer: (n - m) / 5
(iii) (w3 - v3 + x3 + 3wvx) / (w2 + v2 + x2 - 2wv - 2vx + 2wx)
Let y = -v. Numerator: w3 + y3 + x3 - 3wyx = (w + y + x)(w2 + y2 + x2 - wy - yx - wx)
= (w - v + x)(w2 + v2 + x2 + wv + vx - wx)
Denominator: (w - v + x)2
Answer: (w2 + v2 + x2 + wv + vx - wx) / (w - v + x)
(iv) (4y2 - 20yz + 25z2) / (25z2 - 4y2)
Numerator: (2y - 5z)2 = (5z - 2y)2
Denominator: (5z - 2y)(5z + 2y)
Answer: (5z - 2y) / (5z + 2y)
(v) [(x2 + x - 6)(x2 - 7x + 12)] / [(x2 - 6x + 8)(x2 - 9)]
= [(x + 3)(x - 2)(x - 3)(x - 4)] / [(x - 2)(x - 4)(x - 3)(x + 3)]
All terms cancel out!
Answer: 1
(vi) (p4 - 16) / (p2 - 4p + 4)
= (p2 - 4)(p2 + 4) / (p - 2)2 = (p - 2)(p + 2)(p2 + 4) / (p - 2)2
Answer: (p + 2)(p2 + 4) / (p - 2)
End-of-Chapter Exercises
1. Use suitable identities to find products:
(i) (-3x + 4)2 = (4 - 3x)2 = 16 - 24x + 9x2
(ii) (2s + 7)(2s - 7) = (2s)2 - 72 = 4s2 - 49
(iii) (p2 + 1/2)(p2 - 1/2) = (p2)2 - (1/2)2 = p4 - 1/4
(iv) (2n + 7)(2n - 7) = 4n2 - 49
(v) (s - 2t)(s2 + 2st + 4t2) = s3 - (2t)3 = s3 - 8t3
(vi) ( 1/2 r - 4/2 r )2 = ( 1/2 r - 2r )2 = ( -3/2 r )2 = 9/4 r2
(vii) (-3m + 4k - l)2 = 9m2 + 16k2 + l2 - 24mk - 8kl + 6ml
(viii) (xy - 1/3)3 = x3y3 - 3(xy)2(1/3) + 3(xy)(1/9) - 1/27 = x3y3 - x2y2 + xy/3 - 1/27
(ix) ( 7/2 km - 2/3 )3 = 343/8 k3m3 - 49/2 k2m2 + 14/3 km - 8/27
2. Find values using identities:
(i) 17 × 21 = (19 - 2)(19 + 2) = 192 - 22 = 361 - 4 = 357
(ii) 104 × 96 = (100 + 4)(100 - 4) = 1002 - 42 = 10000 - 16 = 9984
(iii) 24 × 16 = (20 + 4)(20 - 4) = 202 - 42 = 400 - 16 = 384
(iv) 1473 = (150 - 3)3 = 1503 - 3(150)2(3) + 3(150)(3)2 - 33 = 3375000 - 202500 + 4050 - 27 = 3176523
(v) 1993 = (200 - 1)3 = 2003 - 3(200)2(1) + 3(200)(1)2 - 1 = 8000000 - 120000 + 600 - 1 = 7880599
(vi) 1273 = (100 + 27)3 or (130 - 3)3 = 2197000 - 152100 + 3510 - 27 = 2048383
(vii) (-107)3 = -(100 + 7)3 = -(1000000 + 210000 + 14700 + 343) = -1225043
(viii) (-299)3 = -(300 - 1)3 = -(27000000 - 270000 + 900 - 1) = -26730899
3. Factor the following algebraic expressions:
(i) 4y2 + 1 + 1/16y2 = (2y + 1/4y)2
(ii) 9m2 - 1/25n2 = (3m + 1/5n)(3m - 1/5n)
(iii) 27b3 - 1/64b3 = (3b - 1/4b)(9b2 + 3/4 + 1/16b2)
(iv) x2 + 5x/6 + 1/6 = (x + 1/2)(x + 1/3)
(v) 27u3 - 1/125 - 27u2/5 + 9u/25 = (3u - 1/5)3
(vi) 64y3 + 1/125 z3 = (4y + 1/5 z)(16y2 - 4/5 yz + 1/25 z2)
(vii) p3 + 27q3 + r3 - 9pqr = (p + 3q + r)(p2 + 9q2 + r2 - 3pq - 3qr - pr)
(viii) 9m2 - 12m + 4 = (3m - 2)2
(ix) 9x3 - 8/3 y3 + z3/3 + 6xyz = 1/3 [ 27x3 + (-2y)3 + z3 - 3(3x)(-2y)(z) ] = 1/3 (3x - 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz - 3xz)
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy = (2x + 3y + 6z)2
(xi) 27u3 - 1/216 - 9u2/2 + u/4 = (3u - 1/6)3
4. Simplify the following:
(i) (4x2 + 4x + 1) / (4x2 - 1) = (2x + 1)2 / [(2x - 1)(2x + 1)] = (2x + 1) / (2x - 1)
(ii) [9(3a3 - 24b3)] / (9a2 - 36b2) = [27(a3 - 8b3)] / [9(a2 - 4b2)] = [3(a - 2b)(a2 + 2ab + 4b2)] / [(a - 2b)(a + 2b)] = [3(a2 + 2ab + 4b2)] / (a + 2b)
(iii) (s3 + 125t3) / (s2 - 2st - 35t2) = [(s + 5t)(s2 - 5st + 25t2)] / [(s - 7t)(s + 5t)] = (s2 - 5st + 25t2) / (s - 7t)
5. Find possible expressions for length and breadth:
(i) Area = 25a2 - 30ab + 9b2 = (5a - 3b)2
Length = 5a - 3b, Breadth = 5a - 3b
(ii) Area = 36s2 - 49t2 = (6s - 7t)(6s + 7t)
Length = 6s + 7t, Breadth = 6s - 7t
6. Find possible dimensions (length, breadth, height) of cuboids:
(i) Volume = 6a2 - 24b2 = 6(a2 - 4b2) = 6(a - 2b)(a + 2b)
Dimensions: 6, (a - 2b), (a + 2b)
(ii) Volume = 3ps2 - 15ps + 12p = 3p(s2 - 5s + 4) = 3p(s - 1)(s - 4)
Dimensions: 3p, (s - 1), (s - 4)
7. Village playground path area problem:
Playground side = 40m. Outer side = 40 + 2s meters.
Area of path = Outer Area - Inner Area = (40 + 2s)2 - 402
= (1600 + 160s + 4s2) - 1600 = 4s2 + 160s
Answer: 4s(s + 40) sq. metres
8. If a number plus its reciprocal equals 10/3, find the number:
Let the number be x.
x + 1/x = 10/3 ⇒ (x2 + 1) / x = 10/3
3x2 - 10x + 3 = 0
3x2 - 9x - x + 3 = 0 ⇒ 3x(x - 3) - 1(x - 3) = 0
(3x - 1)(x - 3) = 0
Answer: The number is 3 or 1/3.
9. Rectangular pool area is 2x2 + 7x + 3, width is 2x + 1. Find length:
Area = 2x2 + 7x + 3 = 2x2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
Since Area = Length × Width and Width = 2x + 1:
Answer: Length = x + 3 hastas.
10*. If both x - 2 and x - 1/2 are factors of px2 + 5x + r, show that p = r:
Let f(x) = px2 + 5x + r.
Since x - 2 is a factor, f(2) = 0 ⇒ p(2)2 + 5(2) + r = 0 ⇒ 4p + r = -10 (Equation 1)
Since x - 1/2 is a factor, f(1/2) = 0 ⇒ p(1/2)2 + 5(1/2) + r = 0 ⇒ p/4 + 5/2 + r = 0 ⇒ p + 4r = -10 (Equation 2)
Equating Equation 1 and Equation 2: 4p + r = p + 4r ⇒ 3p = 3r ⇒ p = r.
Proved: p = r.
11*. If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 - 3abc = -25:
We know (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
52 = a2 + b2 + c2 + 2(10) ⇒ 25 = a2 + b2 + c2 + 20 ⇒ a2 + b2 + c2 = 5
Now, a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - (ab + bc + ca))
= (5)(5 - 10) = 5 × (-5) = -25.
Proved: a3 + b3 + c3 - 3abc = -25.
12*. Check that n3 - n is always divisible by 6 for all natural numbers n:
Factoring n3 - n = n(n2 - 1) = (n - 1)n(n + 1).
This represents the product of three consecutive integers.
1. Among any three consecutive integers, at least one must be divisible by 2.
2. Exactly one of any three consecutive integers must be divisible by 3.
Since 2 and 3 are coprime, the product is always divisible by 2 × 3 = 6.
Conclusion: n3 - n is always divisible by 6.
13*. Find values:
(i) x3 + y3 - 12xy + 64 when x + y = -4
Rewrite as x3 + y3 + 43 - 3(x)(y)(4)
Using identity a3+b3+c3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ca) with a=x, b=y, c=4:
Since a + b + c = x + y + 4 = -4 + 4 = 0, the entire expression evaluates to 0.
Answer: 0
(ii) x3 - 8y3 - 36xy - 216 when x = 2y + 6
Given x - 2y - 6 = 0.
Rewrite expression as x3 + (-2y)3 + (-6)3 - 3(x)(-2y)(-6)
Since sum of terms x + (-2y) + (-6) = 0, the expression equals 0.
Answer: 0
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