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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