## Worksheet on algebraic identities with answer

Here are some questions solution of algebraic identities

Q.1 Write the following in the expanded form

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y)

(iii) (y-6)(y-9)

Q.2 Using algebraic identity evaluate :

(i) (98)²

(ii) 48×52

(iii) 101 ×103

(iv) 153×153 -47×47

Q.3 If x-\frac{1}{x}=5 ,find the value of the

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Q.4 Find z ,if 9z=79²-61²

Q.5 Show that

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Solution:

Problem: 01

Write the following in the expanded form

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y)

(iii) (y-6)(y-9)

Solution:(i)

Using the identity :  (x+y)²=x²+y²+2xy

(3x+4y)²=(3x)²+(4y)²+2.(3x).(4y)

(3x+4y)²=9x²+16y²+24xy

Solution:(ii)

Using the identity : (x-y)(x+y)=x²-y²

(2x-3y)(2x+3y) =(2x)²-(3y)²

(2x-3y)(2x+3y) =4x²-9y²

Solution:(iii)

Using the identity : (x+a)(x+b)=x²+(a+b)x+ab

(y-6)(y-9)

Here a=-6 and b=-9

(y-6)(y-9) =y²+{-6+(-9)}y+(-6)(-9)

(y-6)(y-9) =y²+{-6-9}y+54

(y-6)(y-9) =y²-15y+54

Problem: 02

Using algebraic identity evaluate :

(i) (98)²

(ii) 48×52

(iii) 101 ×103

(iv) 153×153 -47×47

Solution:(i)

(98)² can be written as (100-2)² which is of form (a-b)²

Using the identity : (a-b)²=a²-2ab+b²

(100-2)²=(100)²-2.100.2.+(2)²

(100-2)²=10000-400+4

(100-2)²=9604

Solution(ii)

48×52 can be written as (50-2)×(50+2) which is of form (a-b)(a+b)

Using the identity : (a-b)(a+b)=a²-b²

(50-2)×(50+2)=(50)²-(2)²

(50-2)×(50+2)=2500-4

(50-2)×(50+2)=2496

Solution(iii)

101 ×103 can be written as (100+1)(100+3) which is of form (x+a)(x+b)

Using the identity : (x+a)(x+b) =x²+(a+b)x+ab

We have a=1 and b=3

(100+1)(100+3)=(100)²+(1+3)(100)+1.3

(100+1)(100+3)=10000+(4)(100)+3

(100+1)(100+3)=10000+400+3

(100+1)(100+3)=10403

Solution(iv)

153×153 -47×47  is of form a.a-b.b=a²-b²

Using identity : a²-b²=(a+b)(a-b)

153×153 -47×47 =(153)²-(47)²

(153)²-(47)²=(153+47)(153-47)

(153)²-(47)²=106×200

(153)²-(47)²=21200

Problem:03

If x-\frac{1}{x}=5 ,find the value of the

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Solution(i)

We have  , x-\frac{1}{x}=5

Squaring on both sides

(x-\frac{1}{x})²=(5)²

Using identity : (x-y)²=x²-2xy+y²

(x)²-2.x.\frac{1}{x}+(\frac{1}{x²})=25

x²-2.+(\frac{1}{x²})=25

x²+\frac{1}{x²}-2=25

x²+\frac{1}{x²}=27

Thus ,value of x²+\frac{1}{x²}=27                  ………………(i)

Solution(ii)

Find the value of (x+\frac{1}{x}

(x+\frac{1}{x})²=x²+2.x.\frac{1}{x}+\frac{1}{x²}

(x+\frac{1}{x})²=x²+2+\frac{1}{x²}                  ………………….(ii)

We have value of x²+\frac{1}{x²}=27  from eq(i)

Substitute in eq(ii)

(x+\frac{1}{x})²=27+2

(x+\frac{1}{x})²=29

x+\frac{1}{x}=√29

Problem: 04 Find z ,if 9z=79²-61²

Solution:

We have , 9z=79²-61²

Using the identity :a²-b²=(a+b)(a-b)

So, 79²-61²=(79+61)(79-61) =(140)(18)

9z= 140.18

z=\frac{140.18}{9}

z= 280

Problem: 05

Show that

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Solution(i) :

(2x-7y)²+28xy

Exapand Using identity : (x-y)²=x²-2xy+y²

=(2x)²-2.(2x).(7y)+(7y)²+28xy

= 4x²-28xy+49y²+28xy

=4x²+49y²+0

=4x²+49y²

Hence proved (2x-7y)²+28xy=4x²+49y²

Solution(ii)

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Using identity : (x-y)(x+y)=x²-y²

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)

(a-b)(a+b)=a²-b²,(b-c)(b+c)=b²-c² and (c-a)(c+a)=c²-a²

=a²-b²+b²-c²+c²-a²

=0

Hence proved  ,(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

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