Ex 2.5 NCERT Solutions Class 9 Chapter 2
The solutions to exercise 2.5 (from class 9th's maths NCERT book) are given below. The exercise has 16 questions which have been explained thoroughly with proper reasoning.
(i) (x + 4) (x + 10)
Using identity (x + a) (x + b) = x2 + (a + b)x +
ab
(x + 4) (x + 10) = x2 +
(4 + 10)x + (4 x 10)
=> x2+ 14x + 40
(ii) (x + 8) (x – 10)
Using identity (x + a) (x + b) = x2+ (a + b)x +
ab
(x + 8) (x – 10) = x2+ [8 +
(-10)]x + [( 8 x (-10)]
=> x2+ [8 -10)]x + [- 80]
=> x2 -2x - 80
(iii) (3x + 4) (3x – 5)
Using identity (x + a) (x + b) = x2+ (a + b)x + ab
(3x + 4) (3x – 5)= (3x)2+ [4 + (-5)](3x) + [(
4 x (-5)]
=
9x2+ [4 -5](3x) + (-20)
=
9x2+ (-1)(3x) + (-20)
=
9x2-3x -20
(iv) ( y2+ 3/2 ) (y2 – 3/2 )
Using identity (x + a) (x + b) = x2+ (a + b)x + ab
( y2+ 3/2 ) [y2 + (-3/2 )] = (y2)2+(3/2-3/2)y2+ [3/2x (-3/2)]
= y4 +0 + (-9/4)
= y4 -9/4
Alternate method:
Using identity (x + a) (x - a) = x2 +a2
( y2+ 3/2 ) (y2 -3/2 ) = (y2)2- (3/2)2
= y4 -9/4
(v) (3 – 2x) (3 + 2x)
Using identity (x + a) (x - a) = x2 +a2
(3 – 2x) (3 + 2x) = (3)2- (2x)2
= 9 - 4x2
(i) 103 × 107
(100 + 3) (100 + 7)
Using identity (x + a) (x + b) = x2+ (a + b)x +
ab
(100 + 3) (100 + 7) = 1002+ (3 + 7)(100) + (3 x 7)
=
10000 + 1000 + 21
=
11021
(ii) 95 × 96
(90 + 5) (90 + 6)
Using identity (x + a) (x + b) = x2+ (a + b)x +
ab
(90 + 5) (90 + 6) = 902+ (5 + 6)(90) + (5 x 6)
=
8100 + 990 + 30
=
9120
Alternate method
(100 - 5) (100 - 4)
= 1002+[ (-5) + (-4)](100) +
[(-5) x (-4)]
=
10000 + (-9)(100) + 20
=
10000 - 900 + 20
=
9120
(iii) 104 × 96
(100 + 4) (100 - 4)
Using identity (x + a) (x - a) = x2- a2
(100 + 4) (100 - 4) = 1002- 42
= 10000 - 16
=
9984
(i) 9x2+6xy + y2
Using identity x2 +2xy +y2= (x + y)2
9x2+6xy + y2 can be written as
= (3x)2+ 2(3x)(y) + y2
= (3x - y)2
(ii) 4y2 -4y + 1
Using identity x2 -2xy +y2 = (x - y)2
4y2 -4y + 1
= (2y)2 -2(2y)(1) + 12
= (2y - 1)2
(iii) x2 - y2/100
Using identity (x + a) (x - a) = x2 - a2
x2 - (y2 /100) = x2 -(y/10)2
=
(x +y/10) (x –y/10)
(i) (x +2y +4z)2
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
(x +2y +4z)2
= x2 +(2y)2 +(4z)2 +2(x)(2y)+2(2y)(4z)+ 2(4z)(x)
= x2 +4y2 2+16z2 +4xy+16yz+ 8zx
(ii) (2x – y + z)2
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
(2x – y + z)2
= (2x)2 +(-y)2 +(z)2 +2(2x)(-y)+2(-y)(z)+ 2(z)(2x)
= 4x2 +y2 +z2 -4xy-2yz+ 4zx
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
(-2x + 3 y + 2z)2
= (-2x)2+(3y)2+(2z)2+2(-2x)(3y)+2(3y)(2z)+ 2(2z)(-2x)
= 4x2+9y2+4z2-12xy+12yz- 8zx
(iv) (3a – 7b – c)2
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
(3a – 7b – c)2
= (3a)2+(-7b)2+(-c)2+2(3a)(-7b)+2(-7b)(-c)+ 2(-c)(3a)
= 9a2+49b2+c2-42ab+14bc-6ca
(v) (–2x + 5y – 3z)2
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
= (–2x + 5y – 3z)2
= (-2x)2+(5y)2+(-3z)2+2(-2x)(5y)+2(5y)(-3z)+ 2(-3z)(-2x)
= 4x2+25y2+9z2-20xy- 15yz +12zx
(vi) [1/4a - (1/2)b + 1]2
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
[(1/4)a - (1/2)b + 1]2
= [(1/4)a]2 +[- (1/2b)]2 +(1)2 +2(1/4a)(- 1/2b)+2(- (1/2)b)(1)+ 2(1)((1/4)a)
= (1/16)a2 + (1/4)b2 2+ 1- (1/4)ab - b+ (1/2)a
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) – 2(3y)(4z) – 2(2x)(4z)
= (2x + 3y+ 4z)2
(ii) 2x2+ y2 + 8z2 – 2√2xy + 4 √2yz – 8xz
Using identity (a + b + c)2 = (a2 +b2 +c2 +2ab+2bc+ 2ca)
2x2+ y2 + 8z2 – 2√2xy + 4 √2yz – 8xz
(√2x)2+ (y)2 + (2√2z)2- {2(√2x)(y)} + {2(y) (2√2z)} - {2(2√2z)(√2x)}
(-√2x+ y + 2√2z)2 or (√2x- y - 2√2z)2
(i) (2x + 1)3
Using identity (a + b)3 = a3+b3+3ab(a+b)
(2x + 1)3= (2x)3+ (1)3+3(2x)(1)(2x
+ 1)
= 8x3+1+(6x)(2x + 1)
=8x3+1+12x2 + 6x
(ii) (2a – 3b)3
Using identity (a + b)3 = a3+b3+3ab(a+b)
(2a – 3b)3= (2a)3+(-3b)3+3(2a)(-3b)(2a – 3b)
= 8a3- 27b3- 18ab(2a – 3b)
= 8a3- 27b3- 36a2b + 54b2a
(iii) [ 3/2x+1]3
Using identity (a + b)3 = a3+b3+3ab(a+b)
[ 3/2x+1]3= (3/2x)3+(1)3+3(1)(3/2x)(3/2x+1)
= (27/8)x3+1+ (9/2x)(3/2x+1)
= (27/8)x3+1+ (27/4)x2+ (9/2)x
(iv) [x-2/3y]3
Using identity (a + b)3 = a3+b3+3ab(a+b)
[ x-2/3y]3
= (x)3-(2/3y)3-3(x)(2/3y)(x-2/3y)
= x3- (8/27)y3-(2xy)(x-2/3y)
= x3- (8/27)y3-2x2y+ (4/3) xy2
(i) (99)3
(100-1)3
Using identity (a - b)3= a3-b3-3ab(a-b)
(100-1)3= (100)3 -1 -3(100)(1)(100-1)
= 1000000 - 1 - 300(100-1)
= 999999 - 30000 +300
=
969999 +300
=
970299
(ii) (102)3
(100+2)3
Using identity (a + b)3 = a3+b3+3ab(a+b)
(100+2)3
= (100)3+ (2)3+3(100)(2)(100+2)
= 1000000 + 8 + 600(100+2)
=
1000008+60000 +1200
=
1061208
(iii) (998)3
(1000-2)3
Using identity (a - b)3= a3-b3-3ab(a-b)
(1000-2)3
= (1000)3-(2)3-3(1000)(2)(1000-2)
=
1000000000 - 8 - 6000000 + 12000
=
994011992
(i) 8a3+b3+12a2b+6ab2
8a3+b3+12a2b+6ab2
=
(2a)3+(b)3+6ab(2a+b)
=(2a)3+(b)3+3(2a)(b)(2a+b)
Using a3+b3+3ab(a+b) = (a + b)3
(2a)3+(b)3+3(2a)(b)(2a+b) = (2a+b)3
(ii) 8a3-b3-12a2b+6ab2
8a3-b3-12a2b+6ab2
=
(2a)3+(-b)3-6ab(2a-b)
=(2a)3-(b)3+3(2a)(-b)(2a-b)
Using a3-b3-3ab(a-b) = (a-b)3
(2a)3- (b)3 - 3(2a)(b)(2a-b) = (2a - b)3
(iii) 27 – 125a3 – 135a + 225a2
27 – 125a3 – 135a + 225a2
= (3)3+(-5a)3-45a(3-5a)
=(3)3-(5a)3-
3(3)(5a)(3-5a)
Using a3-b3-3ab(a-b) = (a-b)3
(3)3+(-5a)3+ 3(3)(-5a)(3-5a) = (3 - 5a)3
(iv)64a3 – 27b3 – 144a2b + 108ab2
64a3 – 27b3 – 144a2b + 108ab2
=
(4a)3 + (-3b)3 – 36ab(4a - 3b)
=
(4a)3 - (3b)3 - 3(4a)(3b)(4a-3b)
Using a3-b3-3ab(a-b) = (a-b)3
(4a)3 - (3b)3 - 3(4a)(3b)(4a-3b)= (4a - 3b)3
(v)27p3 -1216 - 92p2 + 14p
27p3 - 1216 - 92p2 +14p
= (3p)3 -(16)3 -3(3p)(16)(3p - 16)
Using a3-b3-3ab(a-b) = (a-b)3
(3p)3 -(16)3 -3(3p)(16)(3p - 16) = (3p - 16)3
(i)
x3+y3= (x+y) (x2- xy + y2)
RHS
= (x+y) (x2- xy + y2)
= x(x2- xy + y2) +y(x2- xy + y2)
= x3-= x2y+xy2+yx2-xy2+y3
= x3+y3=LHS
Hence proved
(ii) x3-y3= (x-y) (x2 +xy + y2)
RHS
= (x-y) (x2 +xy + y2)
= x(x2 +xy + y2) -y(x2 +xy + y2)
= x3-y3=LHS
Hence proved
(i) 27y3+125z3
27y3+125z3= (3y)3+(5z)3
Using (a + b)3= a3+b3+3ab (a + b)
(3y)3+ (5z)3
= (3y)3+ (5z)3+3(3y) (5z) (3y+5z)
= 27y3+125z3+135y2z+225yz2
(ii) 64m3-343n3
=64m3-343n3= (4m)3-(7n)3
= Using
= (4m)3-(7n)3
= 64m3-343n3-336nm2-588mn2
27x3+y3+z3-9xyz
(3x)3+(y)3+(z)3-3(3x)yz
|
Using identity (a)3+(b)3+ (c)3-3abc=(a +b +c)(a2+b2+c2-ab-bc-ca) |
(3x+y+z)[(3x)2+y2+z2-(3x)(y) -(y)(z)-(z)(3x)]
(3x+y+z)[9x2+y2+z2-3xy -yz-3xz]
P.T. x3+y3+z3- 3xyz = 1/2(x+y+z)[(x-y)2+(y-z)2+(z-x)2]
RHS
1/2(x+y+z)[(x-y)2+(y-z)2+(z-x)2]
1/2(x +y +z)[x2+y2-2xy+y2+z2-2xz+z2+x2-2zx]
1/2(x +y +z)[2x2+2y2+2z2-2xy-2xz-2zx]
½ (2) (x +y +z)[x2+y2+z2-xy-xz-zx]
|
Using identity a3+b3+c3-3abc=(a +b +c)(a2+b2+c2-ab-bc-ca) |
(x +y +z)[x2+y2+z2-xy-xz-zx]
= x3+y3+z3-3xyz
|
Using identity
|
x3+y3+z3-3xyz=(x +y +z)[x2+y2+z2-xy
–yz - zx]
Putting x +y +z=0
x3+y3+z3-3xyz = (0)[x2+y2+z2-xy
–yz - zx]
x3+y3+z3-3xyz=0
x3+y3+z3=3xyz
Hence proved
|
Using identity
then If a +b +c=0 a3+b3+c3=3abc |
(i) (-12)3+ (7)3 + (5)3
a=-12, b=7, c=5
a +b +c=-12+7+5=0
Therefore
(-12)3+ (7)3+(5)3= 3(-12)(7)(5)
= - 1260
(ii) (28)3 + (-15)3 + (-13)3
a=28, b=-15, c=-13
a +b +c=28-15-13=0
Therefore
(28)3 + (-15)3 + (-13)3=
3(28)(-15)(-13)
=
16380
(i) 25a2-35a+12
Splitting the middle term
25a2-(20+15)a+12
25a2-20a-15a+12
5a (5a-4)-3(5a-4)
(5a-3)(5a-4)
a=3/5, 4/5
(ii) 35y2-13y-12
Splitting the middle term
35y2-(28-15)y-12
35y2-28y+15y-12
7y (5y-4) + 3(y-4)
(7y+3)(y - 4)
y=-3/7, 4
(i) 3x2-12x
3x(x-4)
Possible
expressions 3, x, (x-4)
(ii) 12ky2+8ky-20k
4k (3y2+2y-5)
4k [3y2 + (5-3)y - 5]
4k [y (3y+5)-1(3y+5)]
4k(y-1) (3y+5)
Possible
expressions = 4k, 3y + 5 and y – 1.
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