April 10, 2026
No Comments
CHAPTER 4
Exploring Algebraic Identities
Verbatim Questions & Solutions | GANITA MANJARI | Class 9 Maths | Part I
Think and Reflect
Page 71
1. What can you say about a and b if (a + b)² < a² + b²?
2. What can you say about a and b if (a + b)² > a² + b²?
3. When will (a + b)² be equal to a² + b²?
2. What can you say about a and b if (a + b)² > a² + b²?
3. When will (a + b)² be equal to a² + b²?
Answer
1. a and b have opposite signs (one positive, one negative)
2. a and b have the same sign (both positive or both negative)
3. When at least one of a or b is zero
2. a and b have the same sign (both positive or both negative)
3. When at least one of a or b is zero
Page 73
What if we replace b by -b in (a + b)² = a² + 2ab + b²?
Answer
We get (a - b)² = a² - 2ab + b², which is also an identity.
Page 76
Label the squares and rectangles in Fig. 4.4 so that it represents the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Answer
The square of side a+b+c can be partitioned into squares of sides a, b, c and rectangles of dimensions a×b, b×c, c×a (each appearing twice).
Page 78
1. Try to evaluate the following using a suitable identity: (i) 35² (ii) 65² (iii) 85² (iv) 105². Do you observe any interesting pattern?
2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Answer
1. 35² = 1225, 65² = 4225, 85² = 7225, 105² = 11025. Pattern: (a5)² = a(a+1) hundred + 25.
2. The figures represent (a + b)(a - b) = a² - b².
2. The figures represent (a + b)(a - b) = a² - b².
Page 79
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Answer
Yes, as long as the two numbers multiply to the constant term (12). 2 and 5 give product 10, not 12, so it would not form a rectangle. The correct split must have product = constant term.
Page 79
1. Figure out the product of x + 2 and x + 3 using algebra tiles.
2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.
2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.
Answer
1. (x+2)(x+3) = x² + 5x + 6
2. x² + 11x + 30 = (x+5)(x+6)
2. x² + 11x + 30 = (x+5)(x+6)
Page 80
We have seen that (x+3)(x+4) = x² + 7x + 12. Also (x+6)(x+7) = x² + 13x + 42. Generalise the pattern to get an expression for (x+a)(x+b).
Answer
(x+a)(x+b) = x² + (a+b)x + ab
Page 85
We already know that x² - y² = (x - y)(x + y). Further, we have verified that x³ - y³ = (x - y)(x² + xy + y²). Observe that x - y is a common factor of x² - y² and x³ - y³. Do you think x - y is also a factor of x⁴ - y⁴? How about x⁵ - y⁵?
Answer
Yes, x - y is a factor of xⁿ - yⁿ for any natural number n.
Page 87
Try to simplify the following rational expression: \(\frac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \frac{(6s - t)^2}{(\_ + \_)(\_ + \_)}\).
Answer
t² + 2ts - 48s² = (t + 8s)(t - 6s). So the expression simplifies to \(\frac{(6s - t)^2}{(t+8s)(t-6s)} = \frac{-(t-6s)^2}{(t+8s)(t-6s)} = -\frac{t-6s}{t+8s}\).
EXERCISE SET 4.1 (Pages 71-72)
1.
Using the identity (a + b)² = a² + 2ab + b², expand the following:
(i) (7x + 4y)²
(ii) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\)
(iii) (2.5p + 1.5q)²
(iv) \(\left(\frac{3}{4}s + 8t\right)^2\)
(v) \(\left(x + \frac{1}{2y}\right)^2\)
(vi) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\)
(i) (7x + 4y)²
(ii) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\)
(iii) (2.5p + 1.5q)²
(iv) \(\left(\frac{3}{4}s + 8t\right)^2\)
(v) \(\left(x + \frac{1}{2y}\right)^2\)
(vi) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\)
Solution
(i) 49x² + 56xy + 16y²
(ii) \(\frac{49}{25}x^2 + \frac{42}{5}xy + \frac{9}{4}y^2\)
(iii) 6.25p² + 7.5pq + 2.25q²
(iv) \(\frac{9}{16}s^2 + 12st + 64t^2\)
(v) \(x^2 + \frac{x}{y} + \frac{1}{4y^2}\)
(vi) \(\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}\)
(ii) \(\frac{49}{25}x^2 + \frac{42}{5}xy + \frac{9}{4}y^2\)
(iii) 6.25p² + 7.5pq + 2.25q²
(iv) \(\frac{9}{16}s^2 + 12st + 64t^2\)
(v) \(x^2 + \frac{x}{y} + \frac{1}{4y^2}\)
(vi) \(\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}\)
2.
Using the same identity, find the values of the following:
(i) (64)²
(ii) (105)²
(iii) (205)²
(i) (64)²
(ii) (105)²
(iii) (205)²
Solution
(i) (60+4)² = 3600 + 480 + 16 = 4096
(ii) (100+5)² = 10000 + 1000 + 25 = 11025
(iii) (200+5)² = 40000 + 2000 + 25 = 42025
(ii) (100+5)² = 10000 + 1000 + 25 = 11025
(iii) (200+5)² = 40000 + 2000 + 25 = 42025
EXERCISE SET 4.2 (Pages 74-75)
1.
Factor completely:
(i) 9x² + 24xy + 16y²
(ii) 4s² + 20st + 25t²
(iii) 49x² + 28xy + 4y²
(iv) \(\frac{64}{9}p^2 + \frac{32}{3}pq + 4q^2\)
*(v) \(\frac{3}{4}a^2 + 4ab + \frac{16}{3}b^2\)
*(vi) \(\frac{9}{5}s^2 + 6sv + 5v^2\)
(i) 9x² + 24xy + 16y²
(ii) 4s² + 20st + 25t²
(iii) 49x² + 28xy + 4y²
(iv) \(\frac{64}{9}p^2 + \frac{32}{3}pq + 4q^2\)
*(v) \(\frac{3}{4}a^2 + 4ab + \frac{16}{3}b^2\)
*(vi) \(\frac{9}{5}s^2 + 6sv + 5v^2\)
Solution
(i) (3x + 4y)²
(ii) (2s + 5t)²
(iii) (7x + 2y)²
(iv) \(\left(\frac{8}{3}p + 2q\right)^2\)
*(v) \(\frac{1}{12}(9a^2 + 48ab + 64b^2) = \frac{1}{12}(3a + 8b)^2\)
*(vi) \(\frac{1}{5}(9s^2 + 30sv + 25v^2) = \frac{1}{5}(3s + 5v)^2\)
(ii) (2s + 5t)²
(iii) (7x + 2y)²
(iv) \(\left(\frac{8}{3}p + 2q\right)^2\)
*(v) \(\frac{1}{12}(9a^2 + 48ab + 64b^2) = \frac{1}{12}(3a + 8b)^2\)
*(vi) \(\frac{1}{5}(9s^2 + 30sv + 25v^2) = \frac{1}{5}(3s + 5v)^2\)
2.
Find the values of the following using the identity (a - b)² = a² - 2ab + b²:
(i) (79)²
(ii) (193)²
(iii) (299)²
(i) (79)²
(ii) (193)²
(iii) (299)²
Solution
(i) (80-1)² = 6400 - 160 + 1 = 6241
(ii) (200-7)² = 40000 - 2800 + 49 = 37249
(iii) (300-1)² = 90000 - 600 + 1 = 89401
(ii) (200-7)² = 40000 - 2800 + 49 = 37249
(iii) (300-1)² = 90000 - 600 + 1 = 89401
EXERCISE SET 4.3 (Pages 76-77)
1.
Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 117²
(ii) 78²
(iii) 198²
(iv) 214²
(v) 1104²
(vi) 1120²
(i) 117²
(ii) 78²
(iii) 198²
(iv) 214²
(v) 1104²
(vi) 1120²
Solution
(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) (1100+4)² = 1210000 + 8800 + 16 = 1218816
(vi) (1100+20)² = 1210000 + 44000 + 400 = 1254400
(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) (1100+4)² = 1210000 + 8800 + 16 = 1218816
(vi) (1100+20)² = 1210000 + 44000 + 400 = 1254400
2.
Factor using suitable identities:
(i) 16y² - 24y + 9
(ii) ???
(iii) \(\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2\)
(iv) ???
(v) 9a² + 4b² + c² - 12ab + 6ac - 4bc
(i) 16y² - 24y + 9
(ii) ???
(iii) \(\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2\)
(iv) ???
(v) 9a² + 4b² + c² - 12ab + 6ac - 4bc
Solution
(i) (4y - 3)²
(iii) \(\left(\frac{m}{3} + 3n + \frac{k}{2}\right)^2\)
(v) (3a - 2b + c)²
(iii) \(\left(\frac{m}{3} + 3n + \frac{k}{2}\right)^2\)
(v) (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
(ii) (3x - 2y + 4z)²
(i) (p + 3q + 7r)²
(ii) (3x - 2y + 4z)²
Solution
(i) p² + 9q² + 49r² + 6pq + 42qr + 14pr
(ii) 9x² + 4y² + 16z² - 12xy - 16yz + 24xz
(ii) 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
Solution
LHS = (a²+b²+c²+2ab-2ac-2bc) + (a²+b²+c²-2ab+2ac-2bc) + (a²+b²+c²-2ab-2ac+2bc) = 3a²+3b²+3c²-2ab-2ac-2bc. This is not equal to 2a²+2b²+2c² in general. So it is NOT an identity.
EXERCISE SET 4.4 (Pages 81-82)
1.
Fill in the blanks to complete the following identities:
(i) s² - 11s + 24 = (________)(________)
(ii) (________)(x + 1) = (3x² - 4x - 7)
(iii) 10x² - 11x - 6 = (2x - ___)(___ + 2)
(iv) 6x² + 7x + 2 = (____________)(___________)
(i) s² - 11s + 24 = (________)(________)
(ii) (________)(x + 1) = (3x² - 4x - 7)
(iii) 10x² - 11x - 6 = (2x - ___)(___ + 2)
(iv) 6x² + 7x + 2 = (____________)(___________)
Solution
(i) (s - 3)(s - 8)
(ii) (3x - 7)(x + 1) = 3x² - 4x - 7, so (3x - 7) is the blank
(iii) 10x² - 11x - 6 = (2x - 3)(5x + 2)
(iv) 6x² + 7x + 2 = (2x + 1)(3x + 2)
(ii) (3x - 7)(x + 1) = 3x² - 4x - 7, so (3x - 7) is the blank
(iii) 10x² - 11x - 6 = (2x - 3)(5x + 2)
(iv) 6x² + 7x + 2 = (2x + 1)(3x + 2)
2.
Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)²
(ii) (27)²
(iii) (23 × 17)
(iv) (135)²
(v) (97)²
(vi) (18 × 29)
(vii) (34 × 43)
(viii) (205)²
(i) (41)²
(ii) (27)²
(iii) (23 × 17)
(iv) (135)²
(v) (97)²
(vi) (18 × 29)
(vii) (34 × 43)
(viii) (205)²
Solution
(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 = 10000+7000+1225=18225
(v) (100-3)² = 10000-600+9=9409
(vi) (18×29) = (18×30 - 18) = 540-18=522
(vii) (34×43) = 34×(40+3)=1360+102=1462
(viii) (200+5)² = 40000+2000+25=42025
(ii) (30-3)² = 900-180+9=729
(iii) (20+3)(20-3) = 400-9=391
(iv) (100+35)² = 10000+7000+1225=18225
(v) (100-3)² = 10000-600+9=9409
(vi) (18×29) = (18×30 - 18) = 540-18=522
(vii) (34×43) = 34×(40+3)=1360+102=1462
(viii) (200+5)² = 40000+2000+25=42025
3.
Factor the following:
(i) 9a² + b² + 4c² - 6ab + 12ac - 4bc
(ii) r² - r - 42
(iii) 64u² + 121v² + 4w² - 176uv - 32uw + 44vw
(i) 9a² + b² + 4c² - 6ab + 12ac - 4bc
(ii) r² - r - 42
(iii) 64u² + 121v² + 4w² - 176uv - 32uw + 44vw
Solution
(i) (3a - b + 2c)²
(ii) (r - 7)(r + 6)
(iii) (8u - 11v - 2w)²
(ii) (r - 7)(r + 6)
(iii) (8u - 11v - 2w)²
Think and Reflect
Page 82
James and Reshma were talking about algebraic identities they learnt in school. According to you, who is correct and why? Try to combine more such identities and find new results.
Answer
Both are correct — they are just different ways of expanding the same expression. James expands directly, while Reshma uses the difference of squares first.
EXERCISE SET 4.5 (Page 87)
1.
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) \(\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}\)
(ii) \(\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}\)
(iii) \(\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}\)
(i) \(\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}\)
(ii) \(\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}\)
(iii) \(\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}\)
Solution
(i) \(\frac{3(p^2 - pq - 6q^2)}{p^2 + 3pq - 10q^2} = \frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}\)
(ii) Using identity w³ + x³ - v³ + 3wvx = (w + x - v)(w² + x² + v² - wx + xv + vw), denominator = (w + x - v)², so expression = \(\frac{w + x - v}{w + x - v} = 1\) (assuming w + x - v ≠ 0)
(iii) \(\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} = 1\)
(ii) Using identity w³ + x³ - v³ + 3wvx = (w + x - v)(w² + x² + v² - wx + xv + vw), denominator = (w + x - v)², so expression = \(\frac{w + x - v}{w + x - v} = 1\) (assuming w + x - v ≠ 0)
(iii) \(\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} = 1\)
END-OF-CHAPTER EXERCISES (Pages 88-90)
1.
Use suitable identities to find the following products:
(i) (-3x + 4)²
(ii) \(\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)\)
(iii) ???
(iv) ???
(v) (s - 2t)(s² + 2st + 4t²)
(vi) ???
(vii) (-3m + 4k - l)²
(viii) ???
(ix) \(\left(\frac{7}{2}k - \frac{2}{3}m\right)^3\)
(i) (-3x + 4)²
(ii) \(\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)\)
(iii) ???
(iv) ???
(v) (s - 2t)(s² + 2st + 4t²)
(vi) ???
(vii) (-3m + 4k - l)²
(viii) ???
(ix) \(\left(\frac{7}{2}k - \frac{2}{3}m\right)^3\)
Solution
(i) 9x² - 24x + 16
(ii) \(p^4 - \frac{1}{4}\)
(v) s³ - 8t³
(vii) 9m² + 16k² + l² - 24mk + 6ml - 8kl
(ix) \(\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3\)
(ii) \(p^4 - \frac{1}{4}\)
(v) s³ - 8t³
(vii) 9m² + 16k² + l² - 24mk + 6ml - 8kl
(ix) \(\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3\)
2.
Find the values using suitable identities:
(i) 17 × 21
(ii) 104 × 96
(iii) 24 × 16
(iv) 147³
(v) 199³
(vi) 127³
(vii) (-107)³
(viii) (-299)³
(i) 17 × 21
(ii) 104 × 96
(iii) 24 × 16
(iv) 147³
(v) 199³
(vi) 127³
(vii) (-107)³
(viii) (-299)³
Solution
(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 - 202500 + 4050 - 27 = 3176523
(v) (200-1)³ = 8000000 - 120000 + 600 - 1 = 7880599
(vi) (125+2)³ = 1953125 + 93750 + 1500 + 8 = 2048383
(vii) (-100-7)³ = -(100+7)³ = -(1000000 + 21000 + 1470 + 343) = -1022813
(viii) (-300+1)³ = -(300-1)³ = -(27000000 - 270000 + 900 - 1) = -26729101
(ii) (100+4)(100-4) = 10000-16=9984
(iii) (20+4)(20-4) = 400-16=384
(iv) (150-3)³ = 3375000 - 202500 + 4050 - 27 = 3176523
(v) (200-1)³ = 8000000 - 120000 + 600 - 1 = 7880599
(vi) (125+2)³ = 1953125 + 93750 + 1500 + 8 = 2048383
(vii) (-100-7)³ = -(100+7)³ = -(1000000 + 21000 + 1470 + 343) = -1022813
(viii) (-300+1)³ = -(300-1)³ = -(27000000 - 270000 + 900 - 1) = -26729101
3.
Factor the following algebraic expressions:
(i) \(4y^2 + 1 + \frac{1}{16y^2}\)
(ii) \(9m^2 - \frac{1}{25n^2}\)
(iii) \(27b^3 - \frac{1}{64b^3}\)
(iv) \(x^2 + \frac{5x}{6} + \frac{1}{6}\)
(v) \(27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}\)
(vi) \(64y^3 + \frac{1}{125}z^3\)
(vii) \(p^3 + 27q^3 + r^3 - 9pqr\)
(viii) \(9m^2 - 12m + 4\)
(ix) \(9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz\)
(x) \(4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy\)
(xi) \(27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}\)
(i) \(4y^2 + 1 + \frac{1}{16y^2}\)
(ii) \(9m^2 - \frac{1}{25n^2}\)
(iii) \(27b^3 - \frac{1}{64b^3}\)
(iv) \(x^2 + \frac{5x}{6} + \frac{1}{6}\)
(v) \(27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}\)
(vi) \(64y^3 + \frac{1}{125}z^3\)
(vii) \(p^3 + 27q^3 + r^3 - 9pqr\)
(viii) \(9m^2 - 12m + 4\)
(ix) \(9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz\)
(x) \(4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy\)
(xi) \(27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}\)
Solution
(i) \(\left(2y + \frac{1}{4y}\right)^2\)
(ii) \(\left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)\)
(iii) \(\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)\)
(iv) \(\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)\)
(v) \(\left(3u - \frac{1}{5}\right)^3\)
(vi) \(\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4y}{5}z + \frac{1}{25}z^2\right)\)
(vii) (p + 3q + r)(p² + 9q² + r² - 3pq - 3qr - pr)
(viii) (3m - 2)²
(ix) Not factorable using simple identities
(x) (2x + 3y + 6z)²
(xi) \(\left(3u - \frac{1}{6}\right)^3\)
(ii) \(\left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)\)
(iii) \(\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)\)
(iv) \(\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)\)
(v) \(\left(3u - \frac{1}{5}\right)^3\)
(vi) \(\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4y}{5}z + \frac{1}{25}z^2\right)\)
(vii) (p + 3q + r)(p² + 9q² + r² - 3pq - 3qr - pr)
(viii) (3m - 2)²
(ix) Not factorable using simple identities
(x) (2x + 3y + 6z)²
(xi) \(\left(3u - \frac{1}{6}\right)^3\)
4.
Simplify the following (assuming denominators ≠ 0):
(i) \(\frac{4x^2 + 4x + 1}{4x^2 - 1}\)
(ii) \(\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}\)
(iii) \(\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}\)
(i) \(\frac{4x^2 + 4x + 1}{4x^2 - 1}\)
(ii) \(\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}\)
(iii) \(\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}\)
Solution
(i) \(\frac{(2x+1)^2}{(2x-1)(2x+1)} = \frac{2x+1}{2x-1}\)
(ii) \(\frac{27(a^3 - 8b^3)}{9(a^2 - 4b^2)} = \frac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)} = \frac{3(a^2+2ab+4b^2)}{a+2b}\)
(iii) \(\frac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)} = \frac{s^2-5st+25t^2}{s-7t}\)
(ii) \(\frac{27(a^3 - 8b^3)}{9(a^2 - 4b^2)} = \frac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)} = \frac{3(a^2+2ab+4b^2)}{a+2b}\)
(iii) \(\frac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)} = \frac{s^2-5st+25t^2}{s-7t}\)
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
(ii) 36s² - 49t²
(i) 25a² - 30ab + 9b²
(ii) 36s² - 49t²
Solution
(i) Area = (5a - 3b)². Possible dimensions: length = 5a - 3b, breadth = 5a - 3b
(ii) Area = (6s - 7t)(6s + 7t). Possible dimensions: length = 6s + 7t, breadth = 6s - 7t
(ii) Area = (6s - 7t)(6s + 7t). Possible dimensions: length = 6s + 7t, breadth = 6s - 7t
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
(ii) 3ps² - 15ps + 12p
(i) 6a² - 24b²
(ii) 3ps² - 15ps + 12p
Solution
(i) 6(a - 2b)(a + 2b). Possible: length = 6, breadth = a - 2b, height = a + 2b
(ii) 3p(s² - 5s + 4) = 3p(s - 1)(s - 4). Possible: length = 3p, breadth = s - 1, height = s - 4
(ii) 3p(s² - 5s + 4) = 3p(s - 1)(s - 4). Possible: length = 3p, breadth = s - 1, height = s - 4
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.
Solution
Area of path = (40 + 2s)² - 40² = 1600 + 160s + 4s² - 1600 = 160s + 4s² = 4s(40 + s)
8.
If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Solution
Let number = x. Then x + 1/x = 10/3 ⇒ 3x² + 3 = 10x ⇒ 3x² - 10x + 3 = 0 ⇒ (3x - 1)(x - 3) = 0 ⇒ x = 3 or x = 1/3
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.
Solution
2x² + 7x + 3 = (2x + 1)(x + 3). So length = x + 3 hastas.
*10.
If both x - 2 and x - 1/2 are factors of px² + 5x + r, show that p = r.
Solution
If x - 2 is a factor, then p(2)² + 5(2) + r = 0 ⇒ 4p + 10 + r = 0 ...(1). If x - 1/2 is a factor, then p(1/4) + 5(1/2) + r = 0 ⇒ p/4 + 5/2 + r = 0 ⇒ p + 10 + 4r = 0 ...(2). Subtracting (1) from (2): (p+10+4r) - (4p+10+r) = 0 ⇒ -3p + 3r = 0 ⇒ p = r.
*11.
If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ - 3abc = -25.
Solution
We know a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). Also a² + b² + c² = (a+b+c)² - 2(ab+bc+ca) = 25 - 20 = 5. So a²+b²+c² - ab - bc - ca = 5 - 10 = -5. Thus a³+b³+c³-3abc = 5 × (-5) = -25.
*12.
By factoring the expression, check that n³ - n is always divisible by 6 for all natural numbers n. Give reasons.
Solution
n³ - n = n(n² - 1) = n(n - 1)(n + 1). This is product of three consecutive integers. Among any three consecutive integers, one is divisible by 3 and at least one is even (divisible by 2). Hence product is divisible by 6.
*13.
Find the value of:
(i) x³ + y³ - 12xy + 64, when x + y = -4
(ii) x³ - 8y³ - 36xy - 216, when x = 2y + 6
(i) x³ + y³ - 12xy + 64, when x + y = -4
(ii) x³ - 8y³ - 36xy - 216, when x = 2y + 6
Solution
(i) Using identity, when x + y = -4, x³ + y³ = (x+y)³ - 3xy(x+y) = -64 - 3xy(-4) = -64 + 12xy. So x³ + y³ - 12xy + 64 = -64 + 12xy - 12xy + 64 = 0
(ii) x - 2y = 6. Using identity, x³ - 8y³ = (x - 2y)³ + 6xy(x - 2y) = 216 + 6xy(6) = 216 + 36xy. So x³ - 8y³ - 36xy - 216 = 0
(ii) x - 2y = 6. Using identity, x³ - 8y³ = (x - 2y)³ + 6xy(x - 2y) = 216 + 6xy(6) = 216 + 36xy. So x³ - 8y³ - 36xy - 216 = 0
Chapter Summary (Pages 90-91)
- Identities are equations that are true for all values of the variables.
- One of the ways to visualise identities is using geometrical models or algebra tiles.
- Identities can also be used to factor algebraic expressions.
- Factorisation of quadratic expressions may be visualised by means of algebra tiles.
- Identities can also be used to simplify calculations such as squaring numbers or evaluating products of numbers.
- Rational algebraic expressions may be simplified by factorisation and removing the common factors.
- Important identities studied: (x+y)² = x² + 2xy + y², (x-y)² = x² - 2xy + y², (x+y+z)² = x² + y² + z² + 2xy + 2yz + 2zx, (x+y)(x-y) = x² - y², (x+a)(x+b) = x² + (a+b)x + ab, x³ - y³ = (x-y)(x² + xy + y²), x³ + y³ = (x+y)(x² - xy + y²), (x+y)³ = x³ + 3x²y + 3xy² + y³, (x-y)³ = x³ - 3x²y + 3xy² - y³, x³ + y³ + z³ - 3xyz = (x+y+z)(x² + y² + z² - xy - xz - yz).
Also Get
Class 9- NCERT- Science Solⁿ.
Class 9-NCERT- English Solⁿ
Class 9-NCERT- Maths
Solⁿ
Class 9-NCERT- Social Science Solⁿ
Class 9 CBSE - SYLLABUS
Study materials
- Refernce Books
- NCERT Solutions
- Syllabus
Send Us A Message
Latest posts
News
Navigating the Shift: A Comprehensive Guide to the Revised NCERT Class 9 Syllabus (2026-27)
Read More »
February 25, 2026
No Comments
Blog
January 24, 2026
No Comments
