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An expression whose left side is equal to the right side is known as equality. Hence, an equality which holds good for all values of variables is called an algebraic identity. Identities give a unique solution for every value of the variable. It should be noted that every identity is an equality but every equality is not considered as an identity. Therefore, (a + b)2=a2 + 2ab+ b2 is considered as identity because this equality is true for all possible values for a and b. On the other hand, let us consider the equation: a2+2ab+ b2=36 cannot be defined as identity as the equation is true for a=5 and b=1 but for other values it does not hold good.
Read Also: Pair of Linear Equations in Two Variables Important Question
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Key Terms: Identity, Expression, Equality, Variables, Binomial, Equation, Algebraic, Trigonometry, Geometry
Standard Algebraic Identities
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There are three different identities which are called standard identities as they are obtained by multiplying a binomial by another binomial. These identities are:
- (a + b)2= a2 + 2ab + b2
- (a - b)2= a2 - 2ab + b2
- (a + b) (a – b) = a2 - b2
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Identity 1: (a + b)2= a2 + 2ab + b2
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This identity is used to calculate the square of the number in a simpler way. In this, the number(n) is divided into two small numbers i.e., a and b which when added gives number n [n= a + b]. So, after applying the identity, the square of the number n can be calculated.
For example,
Calculate the square the 54.
54 can be divided into two numbers i.e., 50 and 4 [54=50 +4].
Here, a=50 and b=4
(54)2= (50+4)2
Applying the equation,
(50+4)2= (50)2 + 2(50)(4) + (4)2
(50+4)2= 2500 + 400 + 16
(50+4)2= 2916
Hence, square of 54 is 2916.
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Identity 2: (a - b)2= a2 - 2ab + b2
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This identity is also used to calculate the square of the number in a simpler way. In this, the number(n) is divided into two small numbers i.e., a and b which when subtracted gives number n [n= a - b]. So, after applying the identity, the square of the number n can be calculated. For example,
Calculate the square the 87.
87 can be divided into two numbers i.e., 90 and 3 [87=90 – 3].
Here, a=90 and b=3
(87)2= (90 - 3)2
Applying the equation,
(90 - 3)2= (90)2 - 2(90)(3) + (3)2
(90 - 3)2= 8100 - 540 + 9
(90 - 3)2= 7569
Hence, square of 87 is 7569.
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Identity 3: (a - b) (a + b) = a2 - b2
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This identity is used to calculate the multiplication of two numbers but that can be expressed into the form (a – b) (a + b). For example,
Calculate 47 × 53
47 can be written as (50 - 3) whereas 53 can be expressed as (50 + 3). So,
47 × 53 = (50 – 3) (50 + 3)
Here, a = 50 and b = 3
Applying the identity
47 × 53 = (50)2 – (3)2
47 × 53 = 2500 – 9
47 × 53 = 2491
Read More: Algebra Formula
Some other useful Identities
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Some of the important identities which can be used for calculation of some complex problems are given below where x, y, a, b and c are the variables.:
Things to Remember
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- An identity is an equality which holds good at every value.
- Every identity is an equality but every equality is not an identity.
- Identities are used to make calculation easier.
- There are three standard identities.
- To prove the equality, one can show LHS = RHS.
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Sample Questions
Question 1: What is the difference between Algebraic Identities and Algebraic Expressions?
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Answer: Algebraic Expression is form of variables and constants without an equal operator whereas Identity is an equality which is true at every value of variable. There are infinite number of expression but only limited numbers of identity. Also, identity also gives the unique answer for every value of the variable. For example, 5a + 6b +35 is an expression because there is no equality and (a + b)2= a2 + 2ab + b2 is considered as an identity.
Question 2: What are the uses of identity?
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Answers: The basic use of the identity is to make the mathematical problem simpler and easier to solve. It reduces the time taken to simplifies it by dividing the large calculation into small parts which when solved gives the correct answer.Also, topics such as geometry, coordinate geometry, trigonometry, calculas have extensive use of the algebraic identities.
Question 3: How many standard identites are there in maths?
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Answer: There are 3 different standard identities which are:
- (a + b)2= a2 + 2ab + b2
- (a - b)2= a2 - 2ab + b2
- (a + b) (a – b) = a2 - b2
Question 4: Proof (a + b)2= a2 + 2ab + b2 is an identity.
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Answer: Let us consider LHS: (a + b)2
The expression (a + b)2 can be written as (a + b) (a + b)
So, (a + b)2 = (a + b) (a + b)
(a + b)2 = a (a + b) + b (a + b)
(a + b)2 = a2 + ab + ba + b2
(a + b)2 = a2 + 2ab + b2 (since ab = ba)
Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication.
Question 5: Use a suitable identity to get each of the following products.
- (x + 3) (x + 3)
- (6y + 5) (6y - 5)
- (2y – 7) (2y – 7)
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Answer: (i) (x + 3) (x + 3)
This can be written as (x + 3)2
Using the identity (a + b)2= a2 + 2ab + b2
Here, a = x and b = 3
(x + 3)2 = x2 + 2(x)(3) + 32
(x + 3)2 = x2 + 6x + 9
So, (x + 3) (x + 3) = x2 + 6x + 9
(ii) (6y + 5) (6y - 5)
This expression Is in form of (a - b) (a + b)
Using the identity (a - b) (a + b) = a2 - b2
Here, a = 6y and b = 5
(6y + 5) (6y - 5) = (6y)2 – (5)2
(6y + 5) (6y - 5) = 36 y2 – 25
Therefore, the answer is 36 y2 – 25.
(iii) (2y – 7) (2y – 7)
This can be written as (2y - 7)2
Using the identity (a - b)2= a2 - 2ab + b2
Here, a = 2y and b = 7
(2y - 7)2 = (2y)2 – 2(2y)(7) + (7)2
(2y - 7)2 = 4y2 – 28y + 49
So, (2y – 7) (2y – 7) = 4y2 – 28y + 49
Question 6: Use the identities to calculate the following:
- (81)2
- (37)2
- 97 × 103
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Answer:(i) (81)2
81 can be expressed as (80 + 1)
So, (81)2 = (80 + 1)2
Using the identity (a + b)2= a2 + 2ab + b2
Here, a = 80 and b = 1
(80 + 1)2 = 802 + 2(80)(1) + 12
(81)2 = 6400 + 160 + 1
So, (81)2 = 6561
(ii) (37)2
This can be written as (40 - 3)2
Using the identity (a - b)2= a2 - 2ab + b2
Here, a = 40 and b = 3
(40 - 3)2 = (40)2 – 2(40)(3) + (3)2
(37)2 = 1600 – 240 + 9
So, (37)2 = 1369
(iii) 97 × 103
This can be expressed as (100 – 3) (100 + 3)
Using the identity (a - b) (a + b) = a2 - b2
Here, a = 100 and b = 3
(100 - 3) (100 + 3) = (100)2 – (3)2
97 × 103 = 10000 – 9
97 × 103 = 9991
Therefore, the answer is 36 y2 – 25.
Question 7: Simplify (2z + 5)2 – (2z - 5)2
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Answer: To expand (2z + 5)2, we will use the identity (a + b)2= a2 + 2ab + b2 and for expanding the (2z - 5)2, the identity (a + b)2= a2 + 2ab + b2 is applied.
Here a = 2z and b = 5
So, (2z + 5)2 = (2z)2 + 2(2z)(5) + 52
(2z + 5)2 = 4z2 + 20z +25
And (2z - 5)2 = (2z)2 - 2(2z)(5) + 52
(2z - 5)2 = 4z2 - 20z +25
Hence, (2z + 5)2 – (2z - 5)2 = 4z2 + 20z +25 – (4z2 - 20z +25)
(2z + 5)2 – (2z - 5)2 = 4z2 + 20z +25 – 4z2 + 20z -25
(2z + 5)2 – (2z - 5)2 = 40z
Therefore, (2z + 5)2 – (2z - 5)2 = 40z
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