Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Identity | Form |
|---|---|
| Square of sum | (a + b)² = a² + 2ab + b² |
| Square of difference | (a − b)² = a² − 2ab + b² |
| Three terms | (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca |
| Difference of squares | a² − b² = (a + b)(a − b) |
| Quadratic | (x + a)(x + b) = x² + (a + b)x + ab |
| Cube of sum / difference | (a ± b)³ = a³ ± 3a²b + 3ab² ± b³ |
| Cubes | a³ ± b³ = (a ± b)(a² ∓ ab + b²) |
| Three cubes | a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) |
Try each problem first, then open the solution to check your working.
Expand (3m − 5n)².
a = 3m, b = 5n: (3m)² − 2(3m)(5n) + (5n)².
= 9m² − 30mn + 25n².
Answer: 9m² − 30mn + 25n²
Expand (2x + y − 3z)².
Squares: 4x², y², 9z².
Cross terms: 2(2x)(y) = 4xy; 2(2x)(−3z) = −12xz; 2(y)(−3z) = −6yz.
Sum: 4x² + y² + 9z² + 4xy − 12xz − 6yz.
Answer: 4x² + y² + 9z² + 4xy − 12xz − 6yz
Find 98² with an identity.
98² = (100 − 2)² = 10000 − 400 + 4 = 9604.
Answer: 9604
Find 103 × 97.
Use: Write as (100 + 3)(100 − 3)
(100 + 3)(100 − 3) = 100² − 3² = 10000 − 9 = 9991.
Answer: 9991
If a + b = 7 and ab = 10, find a² + b² and a³ + b³.
a² + b² = (a + b)² − 2ab = 49 − 20 = 29.
a³ + b³ = (a + b)³ − 3ab(a + b) = 343 − 3(10)(7) = 343 − 210 = 133.
Answer: 29 and 133
If x + 1/x = 5, find x² + 1/x² and x³ + 1/x³.
x² + 1/x² = (x + 1/x)² − 2 = 25 − 2 = 23.
x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 125 − 15 = 110.
Answer: 23 and 110
Factor 25x² − 70xy + 49y².
(5x)² = 25x², (7y)² = 49y² and 2(5x)(7y) = 70xy.
So it is (5x − 7y)².
Answer: (5x − 7y)²
Factor x² − x − 56.
Use: Need a + b = −1 and ab = −56
The pair is −8 and 7: −8 + 7 = −1 and (−8)(7) = −56.
x² − x − 56 = (x − 8)(x + 7).
Answer: (x − 8)(x + 7)
Factor 2x² + x − 6.
Split x using 4 and −3 (product −12, sum 1): 2x² + 4x − 3x − 6.
= 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2).
Answer: (2x − 3)(x + 2)
Factor x³ − 27.
x³ − 3³ = (x − 3)(x² + 3x + 9).
Answer: (x − 3)(x² + 3x + 9)
Factor 8a³ + 125b³.
(2a)³ + (5b)³ = (2a + 5b)(4a² − 10ab + 25b²).
Answer: (2a + 5b)(4a² − 10ab + 25b²)
Expand (2x − 3)³.
(2x)³ − 3(2x)²(3) + 3(2x)(3)² − 3³.
= 8x³ − 36x² + 54x − 27.
Answer: 8x³ − 36x² + 54x − 27
Simplify (x² − 9)/(x² + 5x + 6), x ≠ −2, −3.
Numerator (x − 3)(x + 3); denominator (x + 2)(x + 3).
Cancel x + 3: (x − 3)/(x + 2).
Answer: (x − 3)/(x + 2)
Simplify (a³ − b³)/(a² − b²), a ≠ ±b.
(a − b)(a² + ab + b²)/[(a − b)(a + b)].
= (a² + ab + b²)/(a + b).
Answer: (a² + ab + b²)/(a + b)
The area of a rectangle is 6x² + 17x + 5. Find possible sides.
Split 17x as 2x + 15x (product 30): 6x² + 2x + 15x + 5.
= 2x(3x + 1) + 5(3x + 1) = (3x + 1)(2x + 5).
Answer: 3x + 1 and 2x + 5
A rectangular garden has length 5 m more than its breadth and area 84 m². Find its sides.
Breadth x: x(x + 5) = 84, so x² + 5x − 84 = 0.
(x + 12)(x − 7) = 0; x = 7 (−12 is not possible).
Answer: breadth 7 m, length 12 m
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(i) 49x² + 56xy + 16y².
(ii) 49x²/25 + 2(7x/5)(3y/2) + 9y²/4 = 49x²/25 + 21xy/5 + 9y²/4.
(iii) 6.25p² + 2(2.5)(1.5)pq + 2.25q² = 6.25p² + 7.5pq + 2.25q².
(iv) 9s²/16 + 12st + 64t².
(v) x² + x/y + 1/(4y²).
(vi) 1/x² + 2/(xy) + 1/y².
(i) (60 + 4)² = 3600 + 480 + 16 = 4096.
(ii) (100 + 5)² = 10000 + 1000 + 25 = 11025.
(iii) (200 + 5)² = 40000 + 2000 + 25 = 42025.
(i) (3x)² + 2(3x)(4y) + (4y)² = (3x + 4y)².
(ii) (2s + 5t)².
(iii) (7x + 2y)².
(iv) (8p)² + 2(8p)(2q/3) + (2q/3)² = (8p + 2q/3)².
(v) Take out 3: 3(a² + 4ab/3 + 4b²/9) = 3(a + 2b/3)² = (3a + 2b)²/3.
(vi) Take out 5: 5(9s²/25 + 6sv/5 + v²) = 5(3s/5 + v)² = (3s + 5v)²/5. (Yes, a common factor can be taken out, as in the hint.)
(i) (80 − 1)² = 6400 − 160 + 1 = 6241.
(ii) (200 − 7)² = 40000 − 2800 + 49 = 37249.
(iii) (300 − 1)² = 90000 − 600 + 1 = 89401.
(i) (100 + 17)² = 10000 + 3400 + 289 = 13689.
(ii) (80 − 2)² = 6400 − 320 + 4 = 6084.
(iii) (200 − 2)² = 40000 − 800 + 4 = 39204.
(iv) (200 + 14)² = 40000 + 5600 + 196 = 45796.
(v) (1000 + 100 + 4)² = 1000000 + 10000 + 16 + 200000 + 8000 + 800 = 1218816 (three-term identity).
(vi) (1000 + 100 + 20)² = 1000000 + 10000 + 400 + 200000 + 40000 + 4000 = 1254400 (three-term identity).
(Two-term identities are easiest for (i)–(iv); the three-term identity helps for (v) and (vi).)
(i) (4y)² − 2(4y)(3) + 3² = (4y − 3)².
(ii) (3s/2 + 2t)².
(iii) (m/3 + k/2 + 3n)² (check 2(m/3)(k/2) = mk/3, 2(m/3)(3n) = 2mn, 2(k/2)(3n) = 3nk).
(iv) (p/4 − 4/p)².
(v) (3a − 2b + c)².
(i) p² + 9q² + 49r² + 6pq + 42qr + 14rp.
(ii) 9x² + 4y² + 16z² − 12xy − 16yz + 24zx.
Each square gives a² + b² + c², so the squares add up to 3a² + 3b² + 3c². The cross terms: −2ab − 2ac − 2bc (the others cancel in pairs).
Total: 3a² + 3b² + 3c² − 2ab − 2ac − 2bc ≠ 2a² + 2b² + 2c². Test a = b = c = 1: the left side is 1 + 1 + 1 = 3 and the right side is 6.
No, it is not an identity. (With the fourth square (a + b + c)² added, the sum is 4(a² + b² + c²), which is an identity. This matches Fig. 4.6.)
Take n², (n+1)², (n+2)², (n+3)². First + last − (second + third) = n² + (n+3)² − (n+1)² − (n+2)² = (2n² + 6n + 9) − (2n² + 6n + 5) = 4. E.g. 1, 4, 9, 16: 17 − 13 = 4.
(a + b)² − (a² + b²) = 2ab, so the term 2ab decides.
(i) 2ab < 0: a and b have opposite signs.
(ii) 2ab > 0: a and b have the same sign.
(iii) 2ab = 0: a = 0 or b = 0.
(a + (−b))² = a² + 2a(−b) + (−b)² = a² − 2ab + b², which is the identity for (a − b)².
Along the top and side the lengths are a, b, c. The diagonal squares are a², b², c². The off-diagonal rectangles are ab, ab, bc, bc, ca, ca (each pair is the same rectangle on both sides of the diagonal). Total: a² + b² + c² + 2ab + 2bc + 2ca.
Cut the strip of width b off the right of the a × a square. Move the bottom piece of that strip to the bottom, so that a rectangle of sides (a + b) and (a − b) is formed. The only part left over is the b × b corner. So a² = (a + b)(a − b) + b².
35² = 30 × 40 + 25 = 1225; 65² = 60 × 70 + 25 = 4225; 85² = 80 × 90 + 25 = 7225; 105² = 100 × 110 + 25 = 11025.
Pattern: a number ending in 5, written n5, has square n(n + 1) followed by 25: (10n + 5)² = 100n(n + 1) + 25.
The pieces show the squares of a + b + c, a + b − c, a − b + c and a − b − c. Added together, the cross terms cancel and what remains is 4a² + 4b² + 4c². So the identity is (a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = 4(a² + b² + c²).
One x² tile, 3 x-tiles on one side, 2 x-tiles on the other, and a 2 by 3 array of 6 unit tiles. Total: x² + 5x + 6.
Use one x² tile, 11 x-tiles and 30 unit tiles. Split the x-tiles as 5 + 6 so that the 30 unit tiles fit in a 5 by 6 array. The rectangle has sides x + 5 and x + 6, so x² + 11x + 30 = (x + 5)(x + 6).
No. With 2x + 5x the unit tiles would have to fill a 2 by 5 array = 10 tiles, but we have 12. Only 3 + 4 works because 3 × 4 = 12. Other splits (1 + 6) give a 1 by 6 array = 6 tiles, again not 12.
(x + a)(x + b) = x² + (a + b)x + ab.
(px + a)(qx + b) = px · qx + px · b + a · qx + ab = (pq)x² + (pb + aq)x + ab.
For (2x + 3)(3x + 1): p = 2, a = 3, q = 3, b = 1 gives 6x² + (2 + 9)x + 3 = 6x² + 11x + 3.
Both are correct. James expands (a − b)² first: (a² − 2ab + b²)(a + b) = a³ − a²b − ab² + b³. Reshma groups (a − b)(a + b) = a² − b² and multiplies by (a − b): (a − b)(a² − b²) = a³ − ab² − a²b + b³. Both give the same result.
x⁴ − y⁴ = (x² − y²)(x² + y²) = (x − y)(x + y)(x² + y²), so yes.
x⁵ − y⁵ = (x − y)(x⁴ + x³y + x²y² + xy³ + y⁴): yes. In fact x − y divides xⁿ − yⁿ for every natural n.
Prediction: (x + y)(x² − xy + y²) = x³ + y³.
Denominator: t² + 2ts − 48s² = (t + 8s)(t − 6s) (since 8 − 6 = 2 and 8 × (−6) = −48).
(6s − t)² = (t − 6s)², so the expression is (t − 6s)²/[(t + 8s)(t − 6s)] = (t − 6s)/(t + 8s), with the blanks (t + 8s)(t − 6s).
(i) a + b = −11, ab = 24: −3, −8. (s − 3)(s − 8).
(ii) 3x² − 4x − 7 = 3x² − 7x + 3x − 7 = (3x − 7)(x + 1). The blank is 3x − 7.
(iii) (2x − 3)(5x + 2): check 10x² + 4x − 15x − 6 = 10x² − 11x − 6 ✓.
(iv) 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2).
(i) (40 + 1)² = 1600 + 80 + 1 = 1681.
(ii) (30 − 3)² = 900 − 180 + 9 = 729.
(iii) (20 + 3)(20 − 3) = 400 − 9 = 391.
(iv) (100 + 35)² or n5 rule: 13 × 14 × 100 + 25 = 18225.
(v) (100 − 3)² = 10000 − 600 + 9 = 9409.
(vi) (x + a)(x + b) with x = 20, a = −2, b = 9: 400 + 140 − 18 = 522.
(vii) x = 40, a = −6, b = 3: 1600 − 120 − 18 = 1462.
(viii) (200 + 5)² = 42025.
(i) (3a − b + 2c)².
(ii) (4s − 5t)².
(iii) a + b = −1 and ab = −42: −7 and 6. (r − 7)(r + 6).
(iv) (7g + h)².
(v) (8u − 11v − 2w)² (cross terms: −176uv, −32uw, +44vw ✓).
(i) Numerator 3(p − 3q)(p + 2q); denominator (p + 5q)(p − 2q). No factor is common, so the expression is already simplest: 3(p − 3q)(p + 2q)/[(p + 5q)(p − 2q)].
(ii) Numerator (n − m)³; denominator 5(n − m)². Result (n − m)/5.
(iii) Use a = w, b = −v, c = x in a³ + b³ + c³ − 3abc: numerator = (w − v + x)(w² + v² + x² + wv + vx − wx). Denominator = (w − v + x)². Result (w² + v² + x² + wv + vx − wx)/(w − v + x).
(iv) Numerator (2y − 5z)² = (5z − 2y)²; denominator (5z − 2y)(5z + 2y). Result (5z − 2y)/(5z + 2y).
(v) Numerator (x + 3)(x − 2)(x − 3)(x − 4); denominator (x − 2)(x − 4)(x − 3)(x + 3). Result 1.
(vi) p⁴ − 16 = (p² + 4)(p + 2)(p − 2); denominator (p − 2)². Result (p² + 4)(p + 2)/(p − 2).
(i) 9x² − 24x + 16.
(ii) 4s² − 49.
(iii) p⁴ − 1/4.
(iv) 4n² − 49.
(v) s³ − 8t³.
(vi) 1/(4r²) − 4 + 16r².
(vii) 9m² + 16k² + l² − 24mk + 6ml − 8kl.
(viii) x³ − x²y + xy²/3 − y³/27.
(ix) 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27.
(i) (19 − 2)(19 + 2) = 361 − 4 = 357.
(ii) (100 + 4)(100 − 4) = 10000 − 16 = 9984.
(iii) (20 + 4)(20 − 4) = 400 − 16 = 384.
(iv) (150 − 3)³ = 3375000 − 3(22500)(3) + 3(150)(9) − 27 = 3375000 − 202500 + 4050 − 27 = 3176523.
(v) (200 − 1)³ = 8000000 − 120000 + 600 − 1 = 7880599.
(vi) (100 + 27)³ = 1000000 + 810000 + 218700 + 19683 = 2048383.
(vii) −(100 + 7)³ = −(1000000 + 210000 + 14700 + 343) = −1225043.
(viii) −(300 − 1)³ = −(27000000 − 270000 + 900 − 1) = −26730899.
(i) (2y + 1/(4y))².
(ii) (3m − 1/(5n))(3m + 1/(5n)).
(iii) (3b − 1/(4b))(9b² + 3/4 + 1/(16b²)).
(iv) a + b = 5/6 and ab = 1/6: 1/2 and 1/3. (x + 1/2)(x + 1/3).
(v) (3u − 1/5)³.
(vi) (4y + z/5)(16y² − 4yz/5 + z²/25).
(vii) (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr).
(viii) (3m − 2)².
(ix) Multiply by 3: 27x³ − 8y³ + z³ + 18xyz = (3x)³ + (−2y)³ + z³ − 3(3x)(−2y)(z). So the expression is (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3zx).
(x) As printed, the cross terms are 12xz and 24xy, which do not match any square of a three-term sum. The intended expression is almost certainly 4x² + 9y² + 36z² + 12xy + 24xz + 36yz = (2x + 3y + 6z)² (check: 2(2x)(3y) = 12xy, 2(2x)(6z) = 24xz, 2(3y)(6z) = 36yz). Judgement call: please check your copy; as printed it does not factor over the rationals.
(xi) (3u − 1/6)³.
(i) (2x + 1)²/[(2x − 1)(2x + 1)] = (2x + 1)/(2x − 1).
(ii) 27(a³ − 8b³)/[9(a² − 4b²)] = 3(a − 2b)(a² + 2ab + 4b²)/[(a − 2b)(a + 2b)] = 3(a² + 2ab + 4b²)/(a + 2b).
(iii) (s + 5t)(s² − 5st + 25t²)/[(s − 7t)(s + 5t)] = (s² − 5st + 25t²)/(s − 7t).
(i) = (5a − 3b)(5a − 3b): length = breadth = 5a − 3b (a square).
(ii) = (6s − 7t)(6s + 7t): length 6s + 7t, breadth 6s − 7t.
(i) 6(a² − 4b²) = 6(a − 2b)(a + 2b): dimensions 6, a − 2b, a + 2b.
(ii) 3p(s² − 5s + 4) = 3p(s − 1)(s − 4): dimensions 3p, s − 1, s − 4.
Outer square has side 40 + 2s. Area of path = (40 + 2s)² − 40² = 1600 + 160s + 4s² − 1600 = 4s² + 160s = 4s(s + 40) m².
x + 1/x = 10/3 gives 3x² − 10x + 3 = 0. Split −10x = −9x − x: 3x(x − 3) − (x − 3) = (3x − 1)(x − 3) = 0.
x = 3 or x = 1/3.
2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
Length = x + 3 hastas.
Since px² + 5x + r has degree 2 with leading coefficient p, px² + 5x + r = p(x − 2)(x − 1/2) = px² − (5p/2)x + p.
Comparing constant terms, r = p. (Comparing the x terms, 5 = −5p/2, so p = r = −2.)
a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) = 25 − 20 = 5.
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 5(5 − 10) = −25.
n³ − n = n(n² − 1) = (n − 1) n (n + 1): three consecutive integers.
Among three consecutive integers at least one is even and exactly one is a multiple of 3. So the product is divisible by 2 and by 3, hence by 6 (2 and 3 are co-prime).
(i) Take a = x, b = y, c = 4: a³ + b³ + c³ − 3abc = x³ + y³ + 64 − 12xy. This equals (x + y + 4)(…) = 0 since x + y + 4 = 0. The value is 0.
(ii) Take a = x, b = −2y, c = −6: a³ + b³ + c³ − 3abc = x³ − 8y³ − 216 − 36xy. It equals (x − 2y − 6)(…) = 0 since x = 2y + 6. The value is 0.