**Identities**

Let us examine some products of algebraic expressions more closely.

Consider the
following:

(

*x*+ 9) × (*x*+ 7)
Expanding
the above expression, we get

(

*x*+ 9) × (*x*+ 7) =*x*× (*x*+ 7) +9 × (*x*+ 7)
=

*x*×*x*+*x*× 7 + 9 ×*x*+ 9 × 7 =*x*^{2}+ 7*x*+ 9*x*+ 63
=

*x*^{2}+ 16*x*+ 63
Hence, (

*x*+ 9) (*x*+ 7) =*x*^{2}+ 16*x*+ 63
We find that the above
equality is true for all the values of the variable

*x*.
Let us check the equality
when

*x*= 1
LHS = (1 + 9) (1 +
7) = 10 × 8 = 80

RHS = (1)

^{2}+ 16 × 1 + 63 = 1 + 16 + 63 = 80
Therefore, LHS = RHS

Now, let us check the
equality for

*x*= –2
LHS = (–2 + 9) (–2
+ 7) = 7 × 5 = 35

RHS = (–2)

^{2}+ 16 × (–2) + 63 = 4 – 32 + 63 = 35
Therefore, LHS = RHS

You can give other values to

*x*and verify LHS = RHS.
Such equations
are called

**identities**.##
**Standard
Identities**

Let us learn some useful
identities, which involve the product of two binomials.

###
**Proof of Identity
1**

(

*a*+*b*) × (*a*+*b*)
(

*a*+*b*) (*a*+*b*) =*a*(*a*+*b*) +*b*(*a*+*b*)
=

*a*×*a*+*a*×*b*+*b*×*a*+*b*×*b*=*a*^{2}+*ab*+*ba*+*b*^{2}=*a*^{2}+ 2*ab*+*b*^{2}
Thus,

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

^{}###
**Proof of Identity
2**

(

*a*–*b*) × (*a*–*b*)
(

*a*–*b*) (*a*–*b*) =*a*(*a*–*b*) –*b*(*a*–*b*)
=

*a*×*a*–*a*×*b*–*b*×*a*+*b*×*b*=*a*^{2}–*ab*–*ba*+*b*^{2}=*a*^{2}– 2*ab*+*b*^{2}
Thus,

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

^{}###
**Proof of Identity
3**

(

*a*–*b*) × (*a*+*b*)
(

*a*–*b*) (*a*+*b*) =*a*× (*a*+*b*) –*b*× (*a*+*b*) =*a*×*a*+*a*×*b*+ (–*b*) ×*a*+ (–*b*) ×*b*
=

*a*^{2}+*ab*–*ba*–*b*^{2}=*a*^{2}–*b*^{2}
Thus,

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

^{}###
**Proof of Identity
4**

(

*x*+*a*) (*x*+*b*)
(

*x*+*a*) (*x*+*b*) =*x*× (*x*+*b*) +*a*× (*x*+*b*)
=

*x*×*x*+*x*×*b*+*a*×*x*+*a*×*b*=*x*^{2}+*xb*+*ax*+*ab*=*x*^{2}+ (*b*+*a*)*x*+*ab*
=

*x*^{2}+ (*a*+*b*)*x*+*ab*
Thus,

**(***x***+***a***) (***x***+***b***) =***x*^{2}**+ (***a***+***b***)***x***+***ab*##
**Geometrical
Representation of Identities**

We can
visualize the geometrical representations of the standard identities that we
discussed above.

###
**Geometrical
Representation of ****Identity 1**

(

*a*+*b*)^{2}can be shown as follows.
What is the
product (

*a*+*b*) × (*a*+*b*)?
Area of P =

*a*^{2}
Area of Q =

*ba*
Area of R =

*ab*
Area of S =

*b*^{2}
From the figure given above, (

*a*+*b*)^{2}is the area of the large square, which is the sum of the two smaller squares (P and S) and two smaller rectangles (Q and R). Thus, (*a*+*b*)^{2}is given by the sum of the areas of the regions P, Q, R and S.
(

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

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

This gives
the geometrical proof of the first standard identity.

###
**Geometrical
Representation of ****Identity 2**

In
the following figure, there is a large square whose area is given by

*a*2 and a smaller square within it whose area is given by*b*^{2}.
Extending the line GE, we get GH
and extending the line FE, we get FI. Now, we have two squares and two
rectangles.

Now, area of
AHEI is given by (

*a*–*b*)^{2}.
Area of AHEI = Area of ABCD –
Area of DGEI – Area of GCFE – Area of EFBH

∴ (

*a*–*b*)^{2}=*a*^{2}–*b*(*a*–*b*) –*b*^{2}–*b*(*a*–*b*)
=

*a*^{2}–*ab*+*b*^{2}–*b*^{2}–*ab*+*b*^{2}
=

*a*^{2}–*ab*–*ab*+*b*^{2}–*b*^{2}+*b*^{2}(grouping like terms)
=

*a*^{2}– 2*ab*+*b*^{2}
Thus,
(

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

This gives
the geometrical proof of the second standard identity.

###
**Geometrical
Representation of ****Identity 3**

Let us consider
the geometrical figure of identity 2 once again. There is a large square whose
area is given by

*a*^{2}from which a smaller square whose area is given by*b*^{2}has been removed.
The area of the shaded portion is given by area of larger square – area of smaller square =

*a*^{2}–*b*^{2}.*a*

^{2}–

*b*

^{2}= Area of GDAH + Area of EFBH

=

*a*(*a*–*b*) +*b*(*a*–*b*)
= (

*a*+*b*)(*a*–*b*)
Thus,

*a*^{2}–*b*^{2}= (*a*+*b*)(*a*–*b*)
This
gives the geometrical proof of the third standard identity.

###
**Geometrical
Representation of ****Identity 4**

We know that the equation (

*x*+*a*)(*x*+*b*) =*x*2 + (*a*+*b*)*x*+*ab*is also an**identity**. This relationship can also be**shown geometrically**as follows.
In the
figure, four rectangles A, B, C and D are joined together to form a large
rectangle.

From the
above figure,

1. Area of rectangle A =

*x*^{2}
3. Area of rectangle C =

*xb*
4. Area of rectangle D =

*ab*

Area of
the large rectangle = (

*x*+*a*) (*x*+*b*)
Since the area of the large rectangle

= Area of A + Area of B + Area of C + Area of D

Therefore, (

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

This gives the geometrical proof of the
fourth standard identity.

##
**Solved Examples on Identities**

**Example 1:**Simplify the following:

**a.**(4

*x*+ 3

*y*)

^{2}

**b.**(0.2

*x*+ 0.9

*y*)

^{2}

**Solution:**We know that (

*x*+

*y*)

^{2}=

*x*

^{2}+ 2

*xy*+

*y*

^{2}

^{}

**a.**(4

*x*+ 3

*y*)

^{2}= (4

*x*)

^{2}+ 2(4

*x*) (3

*y*) + (3

*y*)

^{2}

= 16

*x*^{2}+ 24*xy*+ 9*y*^{2}**b.**(0.2

*x*+ 0.9

*y*)

^{2}= (0.2

*x*)

^{2}+ 2(0.2

*x*) (0.9

*y*) + (0.9

*y*)

^{2}

= 0.04

*x*^{2}+ 0.36*xy*+ 0.81*y*^{2}**Example 2:**Find the squares of the following:

**a.**(3

*x*– 4

*y*)

^{2}

**b.**(0.6

*x*– 0.9

*y*)

^{2}

**Solution:**We know that (

*x*–

*y*)

^{2}=

*x*

^{2}– 2

*xy*+

*y*

^{2}

^{}

**a.**(3

*x*– 4

*y*)

^{2}= (3

*x*)

^{2}– 2(3

*x*)(4

*y*) + (4

*y*)

^{2}

= 9

*x*^{2}–24*xy*+ 16*y*^{2}**b.**(0.6

*x*– 0.9

*y*)

^{2}= (0.6

*x*)

^{2}– 2(0.6

*x*) (0.9

*y*) + (0.9

*y*)

^{2}

= 0.36

*x*^{2}– 1.08*xy*+ 0.81*y*^{2}**Example 3:**Using the identity (

*x*–

*y*)(

*x*+

*y*) =

*x*

^{2}–

*y*

^{2}, find the following products.

**a.**(2

*x*– 3

*y*) (2

*x*+ 3

*y*)

**b.**(0.1

*x*+ 0.7

*y*) (0.1

*x*– 0.7

*y*)

**Solution:**

**a.**(2

*x*– 3

*y*) (2

*x*+ 3

*y*) = (2

*x*)

^{2}– (3

*y*)

^{2}= 4

*x*

^{2}– 9

*y*

^{2}

**b.**(0.1

*x*+ 0.7

*y*) (0.1

*x*– 0.7

*y*) = (0.1

*x*)

^{2}– (0.7

*y*)

^{2}= 0.01

*x*

^{2}– 0.49

*y*

^{2}

**Example 4:**Without actual multiplication, find the value of the following:

**a.**(105)

^{2}

**b.**(96)

^{2}

**c.**(712)

^{2}– (288)

^{2}

^{}

**Solution:**

**a.**(105)

^{2}= (100 + 5)

^{2}= (100)

^{2}+ 2 × 100 × 5 + 5

^{2 }[Since (

*a*+

*b*)

^{2}=

*a*

^{2}+ 2

*ab*+

*b*

^{2}]

= 10000 + 1000 + 25 = 11025

**b.**(96)

^{2}= (100 – 4)

^{2}= (100)

^{2}– 2 × 100 × 4 + 4

^{2}[Since (

*a*–

*b*)

^{2}=

*a*

^{2}– 2

*ab*+

*b*

^{2}]

= 10000 – 800 + 16 = 9216

**c.**(712)

^{2}– (288)

^{2}= (712 + 288) (712 – 288)

= 1000 × 424 = 424000

**Example 5:**Using identities, evaluate the following:

**a.**155 × 145

**b.**99.8 × 100.2

**c.**(2.9)

^{2}

^{}

**Solution:**

**a.**155 × 145 = (150 + 5) (150 – 5)

= (150)

^{2}– (5)^{2}[Since (*a*+*b*)(*a*–*b*) =*a*^{2}–*b*^{2}]
= 22500 – 25 = 22475

**b.**99.8 × 100.2 = (100 – 0.2) (100 + 0.2) = (100)

^{2}– (0.2)

^{2}= 10000 – 0.04 = 9999.96

**c.**(2.9)

^{2}= (3 – 0.1)

^{2}= (3)

^{2}– 2 × 3 × 0.1 + (0.1)

^{2}

= 9 – 0.6 + 0.01 = 8.41