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}
+ 2*ab*
+ *b*^{2
}

(*a* − *b*)^{2}
≡ *a*^{2}
− 2*ab*
+ *b*^{2}

(a^{2} − b^{2} ) ≡ (*a + b*)(a − b)

(*x *+ a)(*x *+ b) ≡* x*^{2} + (*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} = __p__^{2} + 2(*p*) (2) + 2^{2} = *p*^{2} + 4*p* + 4

(ii) (3 − a)^{2} = __3 ^{2} − 2(3) (a) + a^{2}
= 9 − 6a + a^{2}__

(iii) (6^{2} − *x*^{2}
) = __(6 + x) (6 − x)__

(iv) (*a + b*)^{2}
− (a − b)^{2} = __a__^{2} + 2*ab* + *b*^{2} − (*a*^{2}
− 2*ab* + *b*^{2})

__= a^{2}
+ 2ab + b^{2} − a^{2} + 2ab − b^{2}
__

__= (1 − 1) a^{2}
+ (2 + 2) ab + (+1 −1) b^{2} = 4ab__

(v) (*a + b*)^{2}
= (*a + b*) × __(a + b)__

(vi) (*m + n*)(**m − n**)

(vii) (*m* + ** 7**)

(xiii) (*k*^{2} -
49) = (k+ ** 7 **)(k -

(ix) *m*^{2} − 6*m + *9 = __(m − 3) ^{2}__

(x) (*m −* 10)(*m + *5) = __m__^{2} + (−10 + 5)*m* + (−10)(5) = *m*^{2}
− 5*m* − 50

** **

**Note**

*x *=1 is the only solution
for 7*x *+ 3=10 whereas any value of* x *satisfies (*x*+2)^{2}* *=* x*^{2}* *+ 4*x *+ 4. So 7* x *+ 3=10 is an equation (*x*+2)^{2}* *=* x*^{2}* *+ 4*x
*+ 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 (3*a* +
4*c*)^{2} by using (*a+b*)^{2} identity.

*Solution:*

Comparing
(3*a* + 4*c*)^{2} with (*a* +
*b*)^{2} , we have *a* = 3*a*,*b* =
4*c*

(3*a* + 4*c*)^{2} = (3*a* )^{2} + 2(3*a*)(4*c* ) +
(4*c*)^{2} (replacing *a* and
*b* values)

= 3^{2} *a*^{2} + (2 × 3 × 4)(*a* × *c*) + 4^{2} *c*^{2}

(3*a* + 4*c*)^{2} = 9*a*^{2} + 24*ac* + 16*c*^{2}

** **

__Example 3.9__

Find the
value of 998^{2} 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}
= *a*^{2}
− 2*ab*
+ *b*^{2}

(1000 −
2)^{2} = (1000)^{2} −
2(1000)(2) + (2)^{2}

(998)^{2}
=
1000000 −
4000 +
4 =
996004

** **

**Think**

Which is corrcet? (3*a*)^{2}
is equal to

(i) 3*a*^{2} (ii)
3^{2}*a *(iii) 6*a*^{2} (iv)
9*a*^{2}

**[Answer: (iv) 9 a^{2}]**

**Solution: **(3*a*)^{2} = 3^{2}*a*^{2}
= 9*a*^{2}

** **

__Example 3.10__

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

*Solution:*

We have (3*x* + 5 *y*)(3*x* − 5 *y*)

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

Now (*a* + *b*)(*a* − *b*) = *a*^{2}
− *b*^{2}

(3*x* + 5*y*)(3*x* −
5*y*) = (3*x*)^{2} − (5*y*)^{2 }(replacing *a* and *b* values)^{}

= 3^{2}
*x*^{2} −
5^{2} *y*^{2}

(3*x* + 5*y*)(3*x* −
5*y*) = 9*x*^{2} − 25*y*^{2}

** **

__Example 3.11__

Expand *y*^{2} −16
by using *a*^{2}−*b*^{2} identity

*Solution:*

*y*^{2}* *−16* *can be written as* y*^{2}* *−* *4^{2}

Comparing
it with *a*^{2} −
*b*^{2} , we get *a* = *y*,*b* = 4

Now *a*^{2} − *b*^{2}
= (*a* + *b*)(*a* − *b*)

*y*^{2}* *−* *4^{2}* *=* *(*
y *+* *4)(* y *−* *4)

*y*^{2}* *−* *16* *=* *(*
y *+* *4)(* y *−* *4)

** **

__Example 3.12__

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

*Solution:*

We have (5*x* + 3)(5*x* + 4)

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

We know (*x* + *a*)(*x* + *b*)
=
*x* ^{2} +
(*a* + *b*)*x*
+
*ab* (replacing
*x*, *a* and *b* values)

(5*x* + 3)(5*x*+4) = (5*x*)^{2} +
(3+4) (5*x*) + (3) (4)

= 5^{2} *x* ^{2} + (7)(5*x*) + 12

(5*x* + 3)(5*x* + 4) = 25*x* ^{2} + 35*x* + 12

** **

**Try these**

Expand using appropriate
identities.

(i) (3 *p* + 2*q*)^{2}

(ii) (105)^{2}

(iii) (2*x* − 5*d*)^{2}

(iv) (98)^{2}

(v) ( *y* − 5)( *y*
+ 5)

(vi) (3*x*)^{2} − 5^{2}

(vii) (2*m* + *n*)(2*m* + *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) ****(3 p + 2q)^{2 }**

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

(*a + b*)^{2} = *a*^{2} + 2*ab* +
*b*^{2}

(3*p* + 2*q*)^{2} = (3*p*)^{2} +
2(3*p*) (2*q*) + (2*q*^{2}) = 9*p*^{2} + 12*pq*
+ 4*q*^{2}

**2) ****(105) ^{2} = (100+ 5)^{2}**

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

(*a + b*)^{2} = *a*^{2} + 2*ab* +
*b*^{2}

(100 + 5)^{2} = (100)^{2} + 2 (100) (5) + 5^{2}
= 10000+ 1000 + 25

105^{2} = 11,025

**3.**** (2 x − 5d)^{2}**

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

(*a − b*)^{2} = *a*^{2} − 2*ab* +
*b*^{2}

(2*x* − 5*d*)^{2} = (2*x*)^{2} − 2
(2*x*) (5*d*) + (5*d*)^{2}

= 2^{2}*x*^{2} − 20*xd* + 5^{2}*d*^{2}
= 4*x*^{2} − 20*xd* + 25*d*^{2}

**4.**** (98) ^{2} = (100 − 2)^{2} **

Comparing (100− 2)^{2} with (*a − b*)^{2}
we get

*a* = 100, *b* = 2

(*a* − *b*)^{2} = *a*^{2} − 2*ab*
+ *b*^{2}

(100 − 2)^{2} = 100^{2} − 2 (100) (2) + 2^{2}

= 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*) = *a*^{2} − *b*^{2}

(*y* − 5)(*y* + 5) = *y*^{2 }−^{ }5^{2}
= *y*^{2} – 25

**6.**** (3 x)^{2} − 5^{2}**

Comparing (3*x*)^{2} − 5^{2} with *a*^{2}
− *b*^{2} we have

*a* = 3*x* ; *b* = 5

(*a*^{2 }−^{ }*b*^{2}) = (*a
+ b*) (*a − b*)

(3*x*)^{2} − 5^{2} = (3*x* + 5) (3*x
− *5) = 3*x* (3*x* − 5) + 5 (3*x* − 5)

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

= 9*x*^{2} − 15*x* + 15*x* − 25 = 9*x*^{2}
– 25

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

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

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

(*x − a*)(*x + b*) = *x*^{2} + (*a + b*)
*x* + *ab*

(2*m* + *n*) (2*m* + *p*) = (2*m*^{2})
+ (*n* + *p*)(2*m*) + (*n*) (*p*)

= 2^{2}*m*^{2} + *n* (2*m*) + *p*
(2*m*) + *np*

= 4*m*^{2} + 2*mn *+ 2*mp* + *np*

**8.**** 203 × 197 = (200 + 3) (200 − 3)**

Comparing (*a + b*) (*a − b*) we have

*a* = 200, *b* = 3

(*a + b*) (*a − b*) = *a*^{2} − *b*^{2}

(200 + 3) (200 − 3) = 200^{2} − 3^{2}

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)

= *x*^{2} − 2*x* − 2*x* + 4

= *x*^{2} − 4*x* + 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) = *y*^{2 }+ (4 + (−3))
*y* + (4) (−3)

= *y*^{2 }+ *y* – 12

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