Identities, Standard Identities, Geometrical Representation of Identities

# Identities, Standard Identities, Geometrical Representation of Identities

## 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 = x2 + 7x + 9x + 63
= x2 + 16x + 63
Hence, (x + 9) (x + 7) = x2 + 16x + 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 = a2 + ab + ba + b2 = a2 + 2ab + b2
Thus, (a + b)2 = a2 + 2ab + b2

### 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 = a2ab ba + b2 = a2 – 2ab + b2
Thus, (a b)2 = a2 – 2ab + b2

### 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
= a2 + ab ba b2 = a2b2
Thus, (a b) (a + b) = a2b2

### 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 = x2 + xb + ax + ab = x2 + (b + a) x + ab
= x2 + (a + b) x + ab
Thus, (x + a) (x + b) = x2 + (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 = a2

Area of Q = ba

Area of R = ab

Area of S = b2

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 = a2 + ba + ab + b2 = a2 + 2ab + b2
Thus, (a + b)2 = a2 + 2ab + b2

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 a2 and a smaller square within it whose area is given by b2.

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 = a2 b(a b) – b2 b(a b)
= a2 ab + b2 b2 ab + b2
= a2 ab ab + b2 b2 + b2 (grouping like terms)
= a2 – 2ab + b2
Thus, (a b)2 = a2 – 2ab + b2

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 a2 from which a smaller square whose area is given by b2 has been removed.

The area of the shaded portion is given by area of larger square – area of smaller square = a2b2.
a2b2 = Area of GDAH + Area of EFBH
= a(a b) + b(a b)
= (a + b)(a b)
Thus, a2b2 = (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) = x2 + (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 = x2
2. Area of rectangle B = xa
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) = x2 + xa + xb + ab = x2 + (a + b)x + ab

This gives the geometrical proof of the fourth standard identity.

## Solved Examples on Identities

Example 1: Simplify the following:

a. (4x + 3y)2                             b. (0.2x + 0.9y)2

Solution: We know that (x + y)2 = x2 + 2xy + y2

a. (4x + 3y)2 = (4x)2 + 2(4x) (3y) + (3y)2
= 16x2 + 24xy + 9y2

b. (0.2x + 0.9y)2 = (0.2x)2 + 2(0.2x) (0.9y) + (0.9y)2
= 0.04x2 + 0.36xy + 0.81y2

Example 2: Find the squares of the following:

a. (3x – 4y)2          b. (0.6x – 0.9y)2

Solution: We know that (x y)2 = x2 – 2xy + y2

a. (3x – 4y)2 = (3x)2 – 2(3x)(4y) + (4y)2
= 9x2 –24xy + 16y2

b. (0.6x – 0.9y)2 = (0.6x)2 – 2(0.6x) (0.9y) + (0.9y)2
= 0.36x2 – 1.08xy + 0.81y2

Example 3: Using the identity (x y)(x + y) = x2y2, find the following products.

a. (2x – 3y) (2x + 3y)                 b. (0.1x + 0.7y) (0.1x – 0.7y)

Solution:
a.  (2x – 3y) (2x + 3y) = (2x)2 – (3y)2 = 4x2 – 9y2

b.  (0.1x + 0.7y) (0.1x – 0.7y) = (0.1x)2 – (0.7y)2 = 0.01x2 – 0.49y2

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 + 52        [Since (a + b)2 = a2 + 2ab + b2]
= 10000 + 1000 + 25 = 11025

b. (96)2 = (100 – 4)2 = (100)2 – 2 × 100 × 4 + 42              [Since (a b)2 = a2 – 2ab + b2]
= 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) = a2b2]
= 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