   Algebraic Identities

# Objective:

This topic  gives an overview of;

• Identity
• Standard Identities
• Applying Identities

# What is an Identity?

Consider the equality (a + 1) (a +2) = a2 + 3a + 2. We shall evaluate both sides of this equality for some value of a, say a = 10. For a = 10,

LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132
RHS = a2 + 3a + 2 = 102 + 3 × 10 + 2 = 100 + 30 + 2 = 132

Thus, the values of the two sides of the equality are equal for a = 10.

Let us now take a = –5,

LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2) = (– 4) × (–3) = 12
RHS = a2 + 3a + 2 = (–5)2 + 3 (–5) + 2 = 25 – 15 + 2 = 10 + 2 = 12

Thus, for a = –5, also LHS = RHS.

We shall find that for any value of a, LHS = RHS. Such an equality, true for every value of the variable in it, is called an identity. Thus, (a + 1) (a + 2) = a2 + 3a + 2 is an identity.

An equation is true for only certain values of the variable in it. It is not true for all values of the variable.

For example, consider the equation  a2 + 3a + 2 = 132, It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc.

# Standard Identities

We shall now study three identities which are very useful in our work. These identities are obtained by multiplying a binomial by another binomial.

Let us first consider the product (a + b) (a + b) or (a + b)2

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

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

Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication. One may verify that for any value of a and any value of b, the values of the two sides are equal.

• Next we consider  (a – b)2 = (a – b) (a – b) = a (a – b) – b (a – b) . We have

= a2 – ab – ba + b2 = a2 – 2ab + b2
=(a – b)2 = a2 – 2ab + b2 -------(II)

• Finally, consider (a + b) (a – b). We have (a + b) (a – b) = a (a – b) + b (a – b)

= a2 – ab + ba – b2 = a2 – b2(since ab = ba)
or (a + b) (a – b) = a2 – b2 -----(III)

The identities (I), (II) and (III) are known as standard identities.

# Applying Identities

We shall now see how, for many problems on multiplication of binomial expressions and also of numbers, use of the identities gives a simple alternative method of solving them.

Example:

Using the Identity (I), find (2x + 3y)2

Solution:

(i) (2x + 3y)2 = (2x)2 + 2(2x) (3y) + (3y)[Using the Identity (I)]

= 4x2 + 12xy + 9y2
We may work out (2x + 3y)2 directly.
(2x + 3y)2 = (2x + 3y) (2x + 3y)
= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)
= 4x2 + 6xy + 6 yx + 9y2 (as xy = yx)
= 4x2 + 12xy + 9y2

Using Identity (I) gave us an alternative method of squaring (2x + 3y).You will realise the simplicity of this method even more if you try to square more complicated binomial expressions than (2x + 3y).

Example :

Using Identity (II), find (4p – 3q)2

Solution:

(i) (4p – 3q)2 =(4p)2 – 2 (4p) (3q) + (3q)2[Using the Identity (II)]

= 16p2 – 24pq + 9q2

Example :

Use the Identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following:(i) 501 × 502

Solution:

501 × 502 = (500 + 1) × (500 + 2) = 5002 + (1 + 2) × 500 + 1 × 2

= 250000 + 1500 + 2 = 251502

# Summary

• An identity is an equality, which is true for all values of the variables in the equality.
• On the other hand, an equation is true only for certain values of its variables. An equation is not an
• identity.
• The following are the standard identities:
• (a + b)2= a2+ 2ab + b2--(I)
• (a – b)2= a2– 2ab + b2--(II)
• (a + b) (a – b) = a2– b2--(III)
•  Another useful identity is (x + a) (x + b) = x2+ (a + b) x + ab --(IV)
• The above four identities are useful in carrying out squares and products of algebraic expressions.
• They also allow easy alternative methods to calculate products of numbers and so on.

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