Algebraic Identities

This topic gives an overview of;

- Identity
- Standard Identities
- Applying Identities

Consider the equality (a + 1) (a +2) = a^{2} + 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 = a^{2 }+ 3a + 2 = 10^{2} + 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 = a^{2} + 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) = a^{2} + 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 a^{2} + 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.

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

= a^{2} + 2ab + b^{2} (since ab = ba)

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

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

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

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

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

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

The identities (I), (II) and (III) are known as **standard 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)^{2 }[Using the Identity (I)]

= 4x^{2} + 12xy + 9y^{2}

We may work out (2x + 3y)^{2} directly.

(2x + 3y)^{2 }= (2x + 3y) (2x + 3y)

= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)

= 4x^{2} + 6xy + 6 yx + 9y^{2} (as xy = yx)

= 4x^{2} + 12xy + 9y^{2}

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)]

= 16p^{2} – 24pq + 9q^{2}

**Example :**

Use the Identity (x + a) (x + b) = x^{2 }+ (a + b) x + ab to find the following:(i) 501 × 502

**Solution:**

501 × 502 = (500 + 1) × (500 + 2) = 500^{2} + (1 + 2) × 500 + 1 × 2

= 250000 + 1500 + 2 = 251502

- 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}= a^{2}+ 2ab + b^{2}--(I) - (a – b)
^{2}= a^{2}– 2ab + b^{2}--(II) - (a + b) (a – b) = a
^{2}– b^{2}--(III) - Another useful identity is (x + a) (x + b) = x
^{2}+ (a + b) x + ab --(IV)

- (a + b)
- 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.

**Cite this Simulator:**

Amrita Learning © 2018. All Rights Reserved