Class 9 · MathsNCERT 2026 Revised Syllabus
Chapter 4: Exploring Algebraic Identities (0904CH04)
Ganita Manjari Part I · 25 questions · 6 exercises
Chapter code: 0904ch04 · Book: Ganita Manjari Part I · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 4. Every exercise has a clear, step-by-step answer for chapter code 0904CH04.
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Exercise Set 4.1(2 questions)
1Using the identity (a+b)²=a²+2ab+b², expand the following:(i) (7x+4y)²(ii) (75x+32y)²(iii) (2.5p+1.5q)²(iv) (34s+8t)²(v) (x+12y)²(vi) (1x+1y)²
School Level Solution
(i) (7x+4y)^2 = 49x^2 + 56xy + 16y^2.
(ii) (7x/5+3y/2)^2 = 49x^2/25 + 21xy/5 + 9y^2/4.
(iii) (2.5p+1.5q)^2 = 6.25p^2 + 7.5pq + 2.25q^2.
(iv) (3s/4+8t)^2 = 9s^2/16 + 12st + 64t^2.
(v) (x+1/(2y))^2 = x^2 + x/y + 1/(4y^2).
(vi) (1/x+1/y)^2 = 1/x^2 + 2/(xy) + 1/y^2.
2Using the same identity, find the values of the following:(i) (64)2(ii) (105)2(iii) (205)2
School Level Solution
i) (64)² = (60+4)² = 3600+480+16 = 4096 Ans.ii) (105)² = (100+5)² = 10000+1000+25 = 11025 Ans.iii) (205)² = (200+5)² = 40000+2000+25 = 42025 Ans.
Exercise Set 4.2(2 questions)
1Factor completely:(i) 9x2 + 24xy + 16y2(ii) 4s2 + 20st + 25t2(iii) 49x2 + 28xy + 4y2 32 4(iv) 64 p2 + pq + q 2 3 9 4 2 *(v) 3a 2 + 4ab + b *(vi) s2 + 6sv + 5v2 (Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises(v) and(vi) above?)
School Level Solution
i) 9x²+24xy+16y² = (3x+4y)² Ans.ii) 4s²+20st+25t² = (2s+5t)² Ans.iii) 49x²+28xy+4y² = (7x+2y)² Ans.iv) 649 p² + 323 pq + 49 q² = (8p/3 + 2q/3)² Ans.v) 3a²+4ab+43 b² = 3(a + 2b/3)² Ans.vi) 95 s²+6sv+5v² = 15(3s+5v)² Ans.
2Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.(i) (79)2(ii) (193)2(iii) (299)2
School Level Solution
i) (79)² = (80−1)² = 6400−160+1 = 6241 Ans.ii) (193)² = (200−7)² = 40000−2800+49 = 37249 Ans.iii) (299)² = (300−1)² = 90000−600+1 = 89401 Ans.
Exercise Set 4.3(4 questions)
1Find 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 Exploring Algebraic Identities 77
School Level Solution
i) 117² = (120−3)² = 13689 Ans.ii) 78² = (80−2)² = 6084 Ans.iii) 198² = (200−2)² = 39204 Ans.iv) 214² = (200+14)² = 45796 Ans.v) 1104² = (1100+4)² = 1218816 Ans.vi) 1120² = (1100+20)² = 1254400 Ans.
2Factor using suitable identities: 9 2(i) 16y2 – 24y + 9(ii) s + 6st + 4t 2 m 2 mk k 2 p2 16(iii) + + + 3nk + 2mn + 9n2(iv) − 2 + 9 3 4 16 p(v) 9a 2 + 4 b2 + c 2 − 12ab + 6ac − 4 bc
School Level Solution
(i) 16y^2−24y+9 = (4y−3)^2.
(ii) 9s^2/4 + 6st + 4t^2 = (3s/2+2t)^2.
(iii) m^2/9 + mk/3 + k^2/4 + 3nk + 2mn + 9n^2 = (m/3+k/2+3n)^2.
(iv) p^2/16 − 2 + 16/p^2 = (p/4 − 4/p)^2, for p ≠ 0.
(v) 9a^2+4b^2+c^2−12ab+6ac−4bc = (3a−2b+c)^2.
3Expand 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
School Level Solution
i) (p+3q+7r)²= p²+9q²+49r²+6pq+42qr+14pr= p²+9q²+49r²+6pq+42qr+14pr Ans.ii) (3x−2y+4z)²= 9x²+4y²+16z²−12xy−16yz+24xz= 9x²+4y²+16z²−12xy−16yz+24xz Ans.
4Is this an identity? (a + b − c )2 + (a − b + c )2 + (a − b − c )2 = 2a 2 + 2b2 + 2c 2. In Grade 8, you were introduced to yet another identity, a2 – b2 = (a + b) (a – b). This can be quite useful if it is rewritten as a2 = (a + b) (a – b) + b2. Look at the following figure. Justify the identity a2 = (a + b) (a – b) + b2 for yourself. In 750 CE, this identity was proposed by Śhrīdharāchārya as a method to quickly compute the squares of numbers. For example, 552 = (55 + 5)(55 – 5) + 52 = 60 × 50 + 25 = 3000 + 25 = 3025. Fig. 4.5
School Level Solution
On expanding the left-hand side, we get 3a^2 + 3b^2 + 3c^2 − 2ab − 2ac + 2bc, which is not equal in general to 2a^2+2b^2+2c^2. Hence the first statement is not an identity as written.
The second identity follows immediately from a^2−b^2=(a+b)(a−b): a^2=(a+b)(a−b)+b^2.
Exercise Set 4.4(3 questions)
1Fill 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 = (____________) (___________)
School Level Solution
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 = (3x+2)(2x+1).
2Select 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
School Level Solution
i) 41²=(40+1)²=1681 Ans.ii) 27²=(30−3)²=729 Ans.iii) 23×17=(20+3)(20−3)=400−9=391 Ans.iv) 135²=(100+35)²=18225 Ans.v) 97²=(100−3)²=9409 Ans.vi) 18×29=(23−5)(23+5)=529−25=504 Ans.vii) 34×43=(38−4)(38+4)=1444−16=1428 Ans.viii) 205²=42025 Ans.
3Factor the following:(i) 9a2 + b2 +4c2 – 6ab + 12ac – 4bc(ii) 16s2 + 25t2 – 40st(iii) r2 – r – 42(iv) 49g2 + 14gh + h2(v) 64u2 + 121v2 + 4w2 – 176uv – 32uw + 44vw
School Level Solution
i) 9a²+b²+4c²−6ab+12ac−4bc = (3a−b+2c)².ii) 16s²+25t²−40st = (4s−5t)².iii) r²−r−42 = (r−7)(r+6).iv) 49g²+14gh+h² = (7g+h)².v) 64u²+121v²+4w²−176uv−32uw+44vw = (8u−11v−2w)².
Exercise Set 4.5(1 questions)
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
School Level Solution
(i) 3(p−3q)(p+2q)/((p+5q)(p−2q)).
(ii) (n−m)/5.
(iii) (w^3+x^3−v^3+3wvx)/(w^2+v^2+x^2−2wv−2vx+2wx) = w+x−v.
(iv) (4y^2−20yz+25z^2)/(25z^2−4y^2) = (5z−2y)/(5z+2y).
(v) 1.
(vi) (p+2)(p^2+4)/(p−2).
End-of-Chapter Exercises(13 questions)
1Use suitable identities to find the following products:(i) (−3x + 4)²(ii) (2s + 7)(2s − 7)(iii) (p² + 12)(p² − 12)(iv) (2n + 7)(2n − 7)(v) (s − 2t)(s² + 2st + 4t²)(vi) (12r − 4r)²(vii) (−3m + 4k − l)²(viii) (x − 13y)³(ix) (72k − 23m)³
School Level Solution
(i) (−3x+4)^2 = 9x^2−24x+16.
(ii) (2s+7)(2s−7)=4s^2−49.
(iii) (p^2+1/2)(p^2−1/2)=p^4−1/4.
(iv) (2n+7)(2n−7)=4n^2−49.
(v) (s−2t)(s^2+2st+4t^2)=s^3−8t^3.
(vi) (1/(2r)−4r)^2 = 1/(4r^2) − 4 + 16r^2.
(vii) (−3m+4k−l)^2 = 9m^2+16k^2+l^2−24mk+6ml−8kl.
(viii) (x−y/3)^3 = x^3−x^2y+xy^2/3−y^3/27.
(ix) (7k/2−2m/3)^3 = 343k^3/8 − 49k^2m/2 + 14km^2/3 − 8m^3/27.
2Find 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
School Level Solution
(i) 17×21=(19−2)(19+2)=357.
(ii) 104×96=(100+4)(100−4)=9984.
(iii) 24×16=(20+4)(20−4)=384.
(iv) 147^3 = 3,176,523.
(v) 199^3 = 7,880,599.
(vi) 127^3 = 2,048,383.
(vii) (−107)^3 = −1,225,043.
(viii) (−299)^3 = −26,730,899.
3Factor the following algebraic expressions:(i) 4y² + 1 + 116y²(ii) 9m² − 125n²(iii) 27b³ − 164b³(iv) x² + 5x6 + 16(v) 27u³ − 1125 − 27u²5 + 9u25(vi) 64y³ + 1125z³(vii) p³ + 27q³ + r³ − 9pqr(viii) 9m² − 12m + 4(ix) 9x³ − 83y³ + 13z³ + 6xyz(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy(xi) 27u³ − 1216 − 9u²2 + u4
School Level Solution
(i) 4y² + 1 + 116y²
= (2y)² + 2(2y)14y + (14y)²
= (2y + 14y)² Ans.
(ii) 9m² − 125n²
= (3m)² − (15n)²
= (3m + 15n)(3m − 15n) Ans.
(iii) 27b³ − 164b³ = (3b)³ − (14b)³
= (3b − 14b)(9b² + 3b·14b + 116b²)
= (3b − 14b)(9b² + 34 + 116b²) Ans.
(iv) x² + 5x6 + 16
= x² + x2 + x3 + 16
= (x + 12)(x + 13) Ans.
(v) 27u³ − 1125 − 27u²5 + 9u25
= (3u − 15)³
= (3u − 15)³ Ans.
(vi) 64y³ + 1125z³ = (4y)³ + (z5)³
= (4y + z5)(16y² − 4y·z5 + z²25)
= (4y + z5)(16y² − 4yz5 + z²25) Ans.
(vii) p³ + 27q³ + r³ − 9pqr = p³ + (3q)³ + r³ − 3(p)(3q)(r)
Using a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca):
= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr) Ans.
(viii) 9m² − 12m + 4 = (3m)² − 2(3m)(2) + 2²
= (3m − 2)² Ans.
(ix) 9x³ − 83y³ + 13z³ + 6xyz
Let A = ∛9 x, B = ∛(83) y, C = ∛(13) z. Then ABC = 2xyz, so 3ABC = 6xyz.
A³ − B³ + C³ + 3ABC, and by a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca):
= (∛9 x − ∛(83) y + ∛(13) z)(∛81 x² + ∛(649) y² + ∛(19) z² + ∛24 xy + ∛(89) yz − ∛3 xz) Ans.
(x) 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 Ans.
(xi) 27u³ − 1216 − 9u²2 + u4 = (3u)³ − (16)³ − 3(3u)²(16) + 3(3u)(16)²
= (3u − 16)³ Ans.
4Simplify the following:(i) 4x² + 4x + 14x² − 1(ii) 9(3a³ − 24b³)9a² − 36b²(iii) s³ + 125t³s² − 2st − 35t²Note: Assume that the denominators are not equal to 0.
School Level Solution
(i) (4x^2+4x+1)/(4x^2−1) = (2x+1)^2/[(2x−1)(2x+1)] = (2x+1)/(2x−1), with 2x±1 ≠ 0.
(ii) 9(3a^3−24b^3)/(9a^2−36b^2) = 3(a^2+2ab+4b^2)/(a+2b), subject to the original denominator restrictions.
(iii) (s^3+125t^3)/(s^2−2st−35t^2) = (s^2−5st+25t^2)/(s−7t), subject to the original denominator restrictions.
5Find 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
School Level Solution
i) 25a²−30ab+9b² = (5a−3b)².ii) 36s²−49t² = (6s−7t)(6s+7t).
6Find 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
School Level Solution
i) 6a²−24b² = 6(a−2b)(a+2b).ii) 3ps²−15ps+12p = 3p(s²−5s+4)=3p(s−1)(s−4).
7The 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.
School Level Solution
Area of playground = 40² = 1600 m².Outer square side = 40+2s.Area of outer square = (40+2s)².Area of path = (40+2s)²−40²= 160s+4s² m².
8If a number plus its reciprocal equals 103, find the number.
School Level Solution
Let the number be x.x + 1x = 103⇒ 3x²−10x+3=0⇒ (3x−1)(x−3)=0∴ x=3 or x=13.
9A 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.
School Level Solution
Area = 2x²+7x+3.Width = 2x+1.Length = (2x²+7x+3)(2x+1).Factor numerator: (2x+1)(x+3).∴ Length = x+3 hastas.
★10If both x − 2 and x − 12 are factors of px² + 5x + r, show that p = r.
School Level Solution
Given, x−2 and x−12 are factors of px²+5x+r.
Therefore, the roots are 2 and 12.
∴ polynomial ∝ (x−2)(x−12)
= x² − 52x + 1
Comparing with px²+5x+r,
−5p2 = 5
p = −2
Also r=p×1=−2
∴ p=r proved.
★11If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.
School Level Solution
Given, a+b+c=5
and ab+bc+ca=10.
Now,
a²+b²+c²=(a+b+c)²−2(ab+bc+ca)
=25−20=5
Using a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)
=(5)(5−10)
=5(−5)=−25 Ans.
★12By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.
School Level Solution
n³−n
=n(n²−1)
=n(n−1)(n+1)
Here, n−1, n and n+1 are three consecutive natural integers.
Among three consecutive integers, one is divisible by 3 and at least one is even.
∴ their product is divisible by 2×3=6.
n³−n is always divisible by 6.
★13Find the value of (i) x³ + y³ − 12xy + 64, when x + y = −4(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6
School Level Solution
(i) x³+y³−12xy+64
Given, x+y=−4
x³+y³=(x+y)³−3xy(x+y)
=(−4)³−3xy(−4)
=−64+12xy
x³+y³−12xy+64 = −64+12xy−12xy+64
=0 Ans.
(ii) x³−8y³−36xy−216
Given, x=2y+6
x³−8y³=(x−2y)(x²+2xy+4y²)
=(6)(x²+2xy+4y²)
Putting x=2y+6, the expression becomes
(2y+6)³−8y³−36(2y+6)y−216
=72y²+216y+216−72y²−216y−216
=0 Ans.