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# KC Sinha Mathematics Solution Class 9 Chapter 4 Algebraic identities exercise 4.1

Page 4.7

### Exercise 4.1

Type 1

#### Question 1

Using suitable identities, f‌ind the following products :

(i) (x+3)(x+3)

Sol :

Here we use identity

a=3 , b=3

(x+a)(x+a)=x2+(a+b)x+ab

⇒x2+(3+3)x+3×3

⇒x2+6x+9

(ii) (x-3)(x+5)

Sol :

Using identity:

(x+a)(x+b)=x2+(a+b)x+ab

a=-3 , b=5

⇒x2+(-3+5)x+(-3)(5)

⇒x2+2x-15

(iii) (5x+2)(5x-3)

Sol:

Using identity:

(x+a)(x+b)=x2+(a+b)x+ab

replace x=5x , a=2 , b=-3

⇒(5x)2+(2-3)5x+(2)(-3)

⇒25x2-5x-6

(iv)

Sol :

Using identity :

(x+a)(x+b)=x2+(a+b)x+ab

Replace x=y2 ,

(v) (2x-1)(2x+5)

Sol :

Using identity :

(x+a)(x+b)=x2+(a+b)x+ab

Replace x=2x , a=-1 ,b=5

⇒(2x)2+(-1+5)2x+(-1)(5)

⇒4x2+8x-5

(vi) (5+2x)(5-2x)

Sol :

Using identity :

a2-b2=(a+b)(a-b)

Replace a=5 ,b=2x

⇒52-(2x)2

⇒25-4x2

(vii)

Sol:

Using identity :

a2-b2=(a+b)(a-b)

Replace

(viii)

Sol :

Using identity :

a2-b2=(a+b)(a-b)

Replace

#### Question 2

Find the following product without multiplying directly :

(i) 105×106

Sol :

Hence , we can express 105 and 106 in terms of 105.5 as follows

⇒105×106

⇒(105.5-0.5)(105.5+0.5)

Using identity :

a2-b2=(a+b)(a-b)

Replace a=105.5 ,b=0.5

⇒(105.5)2-(0.5)2

⇒11130.25-0.25

⇒11130

(ii) 103×98

Sol :

Hence , we can express 103 and 98 in terms of 100.5 as follows

⇒103×98

⇒(100.5+2.5)(100.5-2.5)

Using identity :

a2-b2=(a+b)(a-b)

Replace a=100.5 ,b=2.5

⇒(100.5)2-(2.5)2

⇒10100.25-6.25

⇒10094

(iii) 97×98

Sol :

Hence , we can express 97 and 98 in terms of 97.5 as follows

⇒97×98

⇒(97.5-0.5)(97.5+0.5)

Using identity :

a2-b2=(a+b)(a-b)

Replace a=97.5 ,b=0.5

⇒(97.5)2-(0.5)2

⇒9506.25-0.25

⇒9506

Type 2

#### Question 3

Write the following in expanded form :

(i) (3a+4b+5c)2

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace a=3a,b=4b ,c=5c

⇒(3a)2+(4b)2+(5c)2+2(3a)(4b)+2(4b)(5c)+2(5c)(3a)

⇒9a2+16b2+25c2+24ab+40bc+30ac

(ii) (4a-2b-3c)2

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace a=4a,b=-2b,c=-3c

⇒(4a)2+(-2b)2+(-3c)2+2(4a)(-2b)+2(-2b)(-3c)+2(-3c)(4a)

⇒16a2+4b2+9c2-16ab+12bc-24ac

(iii) (2x-y+z)2

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace a=2x,b=-y,c=-z

⇒(2x)2+(-y)2+(-z)2+2(2x)(-y)+2(-y)(-z)+2(-z)(2x)

⇒4x2+y2+z2-4xy-2yz+4zx

(iv) (2x+3y+z)2

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace a=2x,b=3y,c=z

⇒(2x)2+(3y)2+(z)2+2(2x)(3y)+2(3y)(z)+2(z)(2x)

⇒4x2+9y2+z2+12xy+6yz+4zx

(v)

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace

(vi) (-2x+5y-3z)2

Sol :

Using identity :

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

Replace a=-2x, b=5y, c=-3z

⇒(-2x)2+(5y)2+(-3z)2+2(-2x)(5y)+2(5y)(-3z)+2(-3z)(-2x)

⇒4x2+25y2+9z2-20xy-30yz+12zx

#### Question 4

Write the following cubes in the expanded form :

(i) (3a+4b)3

Sol :

Using identity:

(x+y)3=x3+y3+3x2y+3xy2

Replace x=3a , y=4b

⇒(3a)3+(4b)3+3(3a)2(4b)+3(3a)(4b)2

⇒27a3+64b3+108a2b+144ab2

(ii) (5p-3q)3

Sol :

Using identity:

(x-y)3=x3-y3-3x2y+3xy2

Replace x=5p , y=3q

⇒(5p)3+(3q)3+3(5p)2(3q)+3(5p)(3q)2

⇒125p3-225p2q+135pq2-27q3

(iii)

Sol :

Using identity:

(x-y)3=x3-y3-3x2y+3xy2

Replace x=3p ,

(iv) (4a-3b)3

Sol :

Using identity:

(x-y)3=x3-y3-3x2y+3xy2

Replace x=3p ,

⇒64a3-27b3-144a2b+108ab2

#### Question 5

Using suitable identities, evaluate each of the following :

(i) (104)3

Sol :

=(100+4)3

Using identity:

(x+y)3=x3+y3+3x2y+3xy2

Replace x=100 , y=4

⇒(100)3+(4)3+3(100)2(4)+3(100)(4)2

⇒1000000+64+3×10000×4+3×100×16

⇒1124864

(ii) (999)3

Sol :

=(1000-1)3

Using identity:

(x-y)3=x3-y3-3x2y+3xy2

Replace x=1000 , y=1

⇒(1000)3-(1)3-3(1000)2(1)+3(1000)(1)2

⇒1000000000-1-3000000+3000

⇒997002999

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