Home | | Maths 8th Std | Identities

# Identities

An identity is an equation satisfied by any value that replaces its variable(s).

Identities

We have studied in the previous class about standard algebraic identities. An identity is an equation satisfied by any value that replaces its variable(s). Now, we shall recollect four known identities, which are,

(a + b)2 a 2 + 2ab + b2

(a b)2 a2 2ab + b2

(a2 − b2 ) ≡ (a + b)(a − b)

(x + a)(x + b) ≡ x2 + (a + b)x + ab

Instead of the symbol , we use = to represent an identity without any confusion.

Try these

Expand the following

(i) ( p + 2)2 =  p2 + 2(p) (2) + 22 =  p2 + 4p + 4

(ii) (3 − a)2 =  32 − 2(3) (a) + a2 = 9 − 6a + a2

(iii) (62x2 ) = (6 + x) (6 − x)

(iv) (a + b)2 − (a − b)2a2 + 2ab + b2 − (a2 − 2ab + b2)

= a2 + 2ab + b2a2 + 2abb2

= (1 − 1) a2 + (2 + 2) ab + (+1 −1) b2 = 4ab

(v) (a + b)2 = (a + b) × (a + b)

(vi) (m + n)(m − n) = m2 − n2

(vii) (m7)2 = m2 + 14m + 49

(xiii) (k2 - 49) = (k+ )(k - )

(ix) m2 − 6m + 9 = (m − 3)2

(x) (m − 10)(m + 5) = m2 + (−10 + 5)m + (−10)(5) = m2 − 5m − 50

Note

x =1 is the only solution for 7x + 3=10 whereas any value of x satisfies (x+2)2 = x2 + 4x + 4. So 7 x + 3=10 is an equation (x+2)2 = x2 + 4x + 4 is an identity. An identity is an equation but vice versa is not true.

Application of Identities

The identities give an alternative method of solving problems on multiplication of algebraic expressions and also of numbers.

Example 3.8

Find the value of (3a + 4c)2 by using (a+b)2 identity.

Solution:

Comparing (3a + 4c)2 with (a + b)2 , we have a = 3a,b = 4c (3a + 4c)2 = (3a )2 + 2(3a)(4c ) + (4c)2 (replacing a and b values)

= 32 a2 + (2 × 3 × 4)(a × c) + 42 c2

(3a + 4c)2 = 9a2 + 24ac + 16c2

Example 3.9

Find the value of 9982 by using (a−b)2 identity.

Solution:

We know, 998 can be expressed as (1000 2)

(998)2 = (1000 2)2 This is in the form of (a b)2 , we get a = 1000, b = 2

Now (a b)2 = a2 2ab + b2

(1000 2)2 = (1000)2 2(1000)(2) + (2)2

(998)2 = 1000000 4000 + 4 = 996004

Think

Which is corrcet? (3a)2 is equal to

(i) 3a2 (ii) 32(iii) 6a2 (iv) 9a2

Solution: (3a)2 = 32a2 = 9a2

Example 3.10

Simplify (3x + 5y)(3x 5 y) by using (a+b) (a−b) identity.

Solution: We have (3x + 5 y)(3x  5 y)

Comparing it with (a + b)(a b) we get a = 3x b = 5y

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

(3x + 5y)(3x 5y) = (3x)2 (5y)2 (replacing a and b values)

= 32 x2 52 y2

(3x + 5y)(3x 5y) = 9x2 25y2

Example 3.11

Expand y2 16 by using a2b2 identity

Solution:

y2 16 can be written as y2 42 Comparing it with a2 b2 , we get a = y,b = 4

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

y2 42 = ( y + 4)( y 4)

y2 16 = ( y + 4)( y 4)

Example 3.12

Simplify (5x + 3)(5x + 4) by using (x+a) (x+b) identity.

Solution: We have (5x + 3)(5x + 4)

Comparing it with (x + a)(x + b) , we get x = 5x and a = 3, b = 4

We know (x + a)(x + b) = x 2 + (a + b)x + ab (replacing x, a and b values)

(5x + 3)(5x+4) = (5x)2 + (3+4) (5x) + (3) (4)

= 52 x 2 + (7)(5x) + 12

(5x + 3)(5x + 4) = 25x 2 + 35x + 12

Try these

Expand using appropriate identities.

(i) (3 p + 2q)2

(ii) (105)2

(iii) (2x 5d)2

(iv) (98)2

(v) ( y 5)( y + 5)

(vi) (3x)2 52

(vii) (2m + n)(2m + p)

(viii) 203 ×197

(ix) Find the area of the square whose side is (x 2) units.

(x) Find the area of the rectangle whose length and breadth are ( y + 4) units and ( y 3) units.

Solution:

1) (3p + 2q)2

Comparing (3p + 2q)2 with (a + b)2, we get a = 3p and b = 2q.

(a + b)2 = a2 + 2ab + b2

(3p + 2q)2 = (3p)2 + 2(3p) (2q) + (2q2) = 9p2 + 12pq + 4q2

2) (105)2 = (100+ 5)2

Comparing (100 + 5)2 with (a + b)2, we get a = 100 and b = 5.

(a + b)2 = a2 + 2ab + b2

(100 + 5)2 = (100)2 + 2 (100) (5) + 52 = 10000+ 1000 + 25

1052 = 11,025

3. (2x − 5d)2

Comparing with (a – b)2, we get a = 2x  b = 5d.

(a − b)2 = a2 − 2ab + b2

(2x − 5d)2 = (2x)2 − 2 (2x) (5d) + (5d)2

= 22x2 − 20xd + 52d2 = 4x2 − 20xd + 25d2

4. (98)2 = (100 − 2)2

Comparing (100− 2)2 with (a − b)2 we get

a = 100, b = 2

(ab)2 = a2 − 2ab + b2

(100 − 2)2 = 1002 − 2 (100) (2) + 22

= 10000 − 400 + 4 = 9600 + 4 = 9604

5. (y − 5) (y + 5)

Comparing (y − 5) (y + 5) with (a − b) (a + b) we get

a = y ; b = 5

(a − b) (a + b) = a2b2

(y − 5)(y + 5) = y2 52 = y2 – 25

6. (3x)2 − 52

Comparing (3x)2 − 52 with a2b2 we have

a = 3x ; b = 5

(a2 b2) = (a + b) (a − b)

(3x)2 − 52 = (3x + 5) (3x − 5) = 3x (3x − 5) + 5 (3x − 5)

= (3x) (3x) − (3x) (5) + 5 (3x) − 5 (5)

= 9x2 − 15x + 15x − 25 = 9x2 – 25

7. (2m + n) (2m + p)

Comparing (2m + n) (2m + p) with (x + a) (x + b) we have

x = 2n ; a = n ; b = p

(x − a)(x + b) = x2 + (a + b) x + ab

(2m + n) (2m + p) = (2m2) + (n + p)(2m) + (n) (p)

= 22m2 + n (2m) + p (2m) + np

= 4m2 + 2mn + 2mp + np

8. 203 × 197 = (200 + 3) (200 − 3)

Comparing (a + b) (a − b) we have

a = 200, b = 3

(a + b) (a − b) =  a2b2

(200 + 3) (200 − 3) = 2002 − 32

203 × 197 = 40000 − 9

203 × 197 = 39991

9. Side of a square = x − 2

Area = Side × Side

= (x − 2) (x − 2) = x (x − 2) −2 (x − 2)

= x(x) + (x) (−2) + (−2)(x) + (−2) (−2)

= x2 − 2x − 2x + 4

= x2 − 4x + 4 units square

10. Length of the rectangle = y + 4

breadth of the rectangle = y − 3

Area of the rectangle = length × breadth

= (y + 4)(y − 3) = y2 + (4 + (−3)) y + (4) (−3)

= y2 + y – 12

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 3 : Algebra : Identities | Algebra | Chapter 3 | 8th Maths