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
= a2
– ab – 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
= a2
– b2
Thus, (a – b) (a + b) = a2 – b2
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 = a2
– b2.
a2
– b2
= Area of GDAH + Area of EFBH
= a(a
–
b)
+ b(a
–
b)
= (a
+
b)(a
–
b)
Thus, a2
– b2
= (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
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) = x2 – y2, 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) = a2 – b2]
= 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