Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Q1: Use suitable identities to find the following products.
(i) (-3x + 4)²
= (-3x)² + 2(-3x)(4) + 4²
= 9x² - 24x + 16
Answer: 9x² - 24x + 16
(ii) (2s + 7)(2s - 7)
= (2s)² - 7²
= 4s² - 49
Answer: 4s² - 49
(iii) (p² + 1/2)(p² - 1/2)
= (p²)² - (1/2)²
= p⁴ - 1/4
Answer: p⁴ - 1/4
(iv) (2n + 7)(2n - 7)
= (2n)² - 7²
= 4n² - 49
Answer: 4n² - 49
(v) (s - 2t)(s² + 2st + 4t²)
Using:
(a - b)(a² + ab + b²) = a³ - b³
= s³ - (2t)³
= s³ - 8t³
Answer: s³ - 8t³
(vi) (1/2r - 4r)²
= (1/2r)² - 2(1/2r)(4r) + (4r)²
= 1/(4r²) - 4 + 16r²
Answer:
16r² - 4 + 1/(4r²)
(vii) (-3m + 4k - l)²
Using:
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
= 9m² + 16k² + l²
-24mk + 6ml - 8kl
Answer:
9m² + 16k² + l² - 24mk + 6ml - 8kl
(viii) (x - y/3)³
Using:
(a-b)³ = a³ - 3a²b + 3ab² - b³
= x³ - x²y + xy²/3 - y³/27
Answer:
x³ - x²y + xy²/3 - y³/27
(ix) (7k/2 - 2m/3)³
= (7k/2)³ - 3(7k/2)²(2m/3) + 3(7k/2)(2m/3)² - (2m/3)³
= 343k³/8 - 49k²m/2 + 14km²/3 - 8m³/27
Answer:
343k³/8 - 49k²m/2 + 14km²/3 - 8m³/27
Q2: Find the values using suitable identities.
(i) 17 × 21
= (19 - 2)(19 + 2)
= 19² - 2²
= 361 - 4
= 357
Answer: 357
(ii) 104 × 96
= (100 + 4)(100 - 4)
= 100² - 4²
= 10000 - 16
= 9984
Answer: 9984
(iii) 24 × 16
= (20 + 4)(20 - 4)
= 20² - 4²
= 400 - 16
= 384
Answer: 384
(iv) 147³
147 = 150 - 3
(147)³ = (150 - 3)³
= 3176523
Answer: 3176523
(v) 199³
199 = 200 - 1
(199)³ = (200 - 1)³
= 7880599
Answer: 7880599
(vi) 127³
127 = 120 + 7
(127)³ = (120 + 7)³
= 2048383
Answer: 2048383
(vii) (-107)³
= -(107³)
= -1225043
Answer: -1225043
(viii) (-299)³
= -(299³)
= -26730899
Answer: -26730899
Q3: Factor the following algebraic expressions.
(i)
4y² + 1 + 1/(16y²)
= (2y + 1/(4y))²
Answer:
(2y + 1/(4y))²
(ii)
9m² - 1/(25n²)
= (3m)² - (1/5n)²
= (3m - 1/5n)(3m + 1/5n)
Answer:
(3m - 1/5n)(3m + 1/5n)
(iii)
27b³ - 1/(64b³)
= (3b)³ - (1/4b)³
= (3b - 1/4b) (9b² + 3/4 + 1/(16b²))
Answer:
(3b - 1/4b)(9b² + 3/4 + 1/(16b²))
(iv)
x² + 5x/6 + 1/6
= (x + 1/2)(x + 1/3)
Answer:
(x + 1/2)(x + 1/3)
(v)
27u³ - 1/125 - 27u²/5 + 9u/25
= (3u - 1/5)³
Answer:
(3u - 1/5)³
(vi)
64y³ + z³/125
= (4y)³ + (z/5)³
= (4y + z/5) (16y² - 4yz/5 + z²/25)
Answer:
(4y + z/5)(16y² - 4yz/5 + z²/25)
(vii)
p³ + 27q³ + r³ - 9pqr
= (p + 3q + r) (p² + 9q² + r² - 3pq - pr - 3qr)
Answer:
(p + 3q + r) (p² + 9q² + r² - 3pq - pr - 3qr)
(viii)
9m² - 12m + 4
= (3m - 2)²
Answer: (3m - 2)²
(ix)
9x³ - (8/3)y³ + z³/3 + 6xyz
= (3x - 2y/3 + z/3) (9x² + 4y²/9 + z²/9 + 2xy - xz + 2yz/9)
Answer: Factored form as above.
(x)
4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= (2x + 3y + 6z)²
Answer:
(2x + 3y + 6z)²
(xi)
27u³ - 1/216 - 9u²/2 + u/4
= (3u - 1/6)³
Answer:
(3u - 1/6)³
Q4: Simplify the following (Assume the denominators are not equal to 0).
(i)
(4x² + 4x + 1)/(4x² - 1)
Factor numerator:
4x² + 4x + 1 = (2x + 1)²
Factor denominator:
4x² - 1 = (2x + 1)(2x - 1)
Therefore,
[(2x + 1)²]/[(2x + 1)(2x - 1)]
= (2x + 1)/(2x - 1)
Answer:
(2x + 1)/(2x - 1)
(ii)
9(3a³ - 24b³)/(9a² - 36b²)
= 27(a³ - 8b³)/9(a² - 4b²)
= 3(a³ - 8b³)/(a² - 4b²)
Factor:
a³ - 8b³ = (a - 2b)(a² + 2ab + 4b²)
a² - 4b² = (a - 2b)(a + 2b)
Cancel common factor:
= 3(a² + 2ab + 4b²)/(a + 2b)
Answer:
3(a² + 2ab + 4b²)/(a + 2b)
(iii)
(s³ + 125t³)/(s² - 2st - 35t²)
Factor numerator:
s³ + 125t³
= (s + 5t)(s² - 5st + 25t²)
Factor denominator:
s² - 2st - 35t²
= (s - 7t)(s + 5t)
Cancel common factor:
= (s² - 5st + 25t²)/(s - 7t)
Answer:
(s² - 5st + 25t²)/(s - 7t)
Q5: Find possible expressions for the length and breadth of each rectangle.
(i) Area = 25a² − 30ab + 9b²
Factor the area:
25a² − 30ab + 9b²
= (5a − 3b)²
Hence, possible dimensions are:
| Length | Breadth |
|---|---|
| 5a − 3b | 5a − 3b |
(ii) Area = 36s² − 49t²
= (6s − 7t)(6s + 7t)
Hence, possible dimensions are:
| Length | Breadth |
|---|---|
| 6s + 7t | 6s − 7t |
Q6: Find possible expressions for the length, breadth and height of each cuboid.
(i) Volume = 6a² − 24ab²
= 6a(a² − 4b²)
= 6a(a − 2b)(a + 2b)
Possible dimensions:
| Length | Breadth | Height |
|---|---|---|
| 6a | a − 2b | a + 2b |
(ii) Volume = 3ps² − 15ps + 12p
= 3p(s² − 5s + 4)
= 3p(s − 1)(s − 4)
Possible dimensions:
| Length | Breadth | Height |
|---|---|---|
| 3p | s − 1 | s − 4 |
Q7: Area of the path around a square playground.
Side of playground = 40 m
Path width = s m
Outer square side:
40 + 2s
Area of outer square:
(40 + 2s)²
= 1600 + 160s + 4s²
Area of playground:
40² = 1600
Area of path:
(1600 + 160s + 4s²) − 1600
= 160s + 4s²
Answer:
4s² + 160s square metres
Q8: If a number plus its reciprocal equals 10/3, find the number.
Let the number be x.
x + 1/x = 10/3
Multiplying by 3x:
3x² + 3 = 10x
3x² − 10x + 3 = 0
Factorising:
(3x − 1)(x − 3) = 0
Therefore,
x = 1/3
or
x = 3
Answer:
3 and 1/3
Q9: Find the length of the rectangular pool.
Area = 2x² + 7x + 3
Width = 2x + 1
Length = Area ÷ Width
Factor the area:
2x² + 7x + 3
= (2x + 1)(x + 3)
Therefore,
Length = x + 3
Answer:
x + 3 hastas
Q10: If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.
Since x − 2 and x − 1/2 are factors,
px² + 5x + r
= k(x − 2)(x − 1/2)
= k(x² − 5x/2 + 1)
Comparing coefficients:
px² + 5x + r
The coefficient of x² and constant term are both multiplied by the same k.
Hence,
p = r
Hence proved.
Q11: If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.
Use the identity:
a³+b³+c³−3abc
= (a+b+c)(a²+b²+c²−ab−bc−ca)
Now,
a²+b²+c²
= (a+b+c)² − 2(ab+bc+ca)
= 25 − 20
= 5
Therefore,
a²+b²+c²−ab−bc−ca
= 5 − 10
= −5
Hence,
a³+b³+c³−3abc
= 5(−5)
= −25
Hence proved.
Q12: Show that n³ − n is divisible by 6.
n³ − n
= n(n² − 1)
= n(n − 1)(n + 1)
These are three consecutive integers.
Among any three consecutive integers:
- One is divisible by 2.
- One is divisible by 3.
Therefore their product is divisible by:
2 × 3 = 6
Hence,
n³ − n
is divisible by 6 for every natural number n.
Q13: Find the value.
(i) x³ + y³ − 12xy + 64, when x + y = −4
Using:
x³ + y³ = (x+y)³ − 3xy(x+y)
= (−4)³ − 3xy(−4)
= −64 + 12xy
Substitute:
(−64 + 12xy) − 12xy + 64
= 0
Answer: 0
(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6
x³ − 8y³ − 216 − 36xy
= (x − 2y − 6) (x² + 4y² + 36 + 2xy + 6x − 12y)
Since:
x = 2y + 6
x − 2y − 6 = 0
Therefore entire expression becomes:
0
Answer: 0
Important identities
Here are some important algebraic identities that are commonly used in mathematics:
Important Algebraic Identities