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Algebraic Identities For Class 9

In the algebraic identities for class 9, we will cover all the identities mentioned in the CBSE class 9 (NCERT) syllabus. These algebraic identities basically carry variable equations in such a way, that the Left-hand side (L.H.S) of the equation is equal to its Right-hand side(R.H.S).

The algebraic identities for class 9 consist of identities of all the algebraic formulas and expressions. You must have learned algebra formulas for class 9, which are mathematical rules expressed in symbols but the algebraic identities represent that the equation is true for all the values of the variables.
For example; (x+1) (x+2) = x2 + 3x + 2.

If we put the value for x=1,

then we get, (1+1) (1+2) = 12 + 3.1 + 2

2 . 3 = 1 + 3 + 2

6 = 6 (L.H.S = R.H.S)

Therefore, from the above example, it is clear that the given equation is an identity. But do you believe that every true equation is an identity equation? Well, the answer is, not every algebraic equation holds the algebraic identity. Say for example, x2 +2x+1 = 110 is an equation but not an identity. Let us prove it by putting the value of x. Let x = 1, then,

12 +2.1+1 = 110

1 + 2 + 1 = 110

4 ≠ 110

Therefore, the definition of algebraic identity does not rely on this algebraic equation, x2 +2x+1 = 110.

Now, let us discuss the important algebraic identities covered in the class 9th syllabus.

Algebraic Identities Class 9 Formulas

Let us consider x, y and z are variables.

Algebraic Identities for Two Variables x and y.

(x + y)2 = x2 + y2 + 2xy
(x – y)2 = x2 + y2 – 2xy
x2 – y2 = (x + y) (x – y)
(x + a) (x + b) = x2 + (a + b)x + ab ; a and b are two constant values
(x + y)3 = x3 + y3 + 3xy(x + y)
(x – y)3 = x3 – y3 – 3xy(x – y)

Algebraic Identities for Three Variables x, y and z.

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Proof of Algebraic Identities

Let us prove here, few identities.

  1. (x + y)2 = x2 + y2 + 2xy

L.H.S. = (x + y)2

L.H.S. = (x + y) (x + y)

By multiplying each term, we get,

L.H.S = x2 + xy + xy + y2

L.H.S. = x2 + 2xy + y2

L.H.S. = R.H.S.

  1. (x – y)2 = x2 + y2 – 2xy

By taking L.H.S.,

(x – y)2 = (x – y) (x – y)

(x – y)2 = x2 – xy – xy + y2

(x – y)2 = x2 – 2xy + y2

L.H.S. = R.H.S. Hence, proved.

  1. x2 – y2 = (x + y) (x – y)

By taking R.H.S and multiplying each term.

(x + y) (x – y) = x2 – xy + xy – y2

(x + y) (x – y) = x2 – y2

Or

x2 – y2 = (x + y) (x – y)

L.H.S. = R.H.S. Hence proved.

In the same way, you can prove the other above given algebraic identities.

Solved Problems on Algebraic Identities

Problem: Solve (x + 3) (x – 3) using algebraic identities.

Solution: By the algebraic identity, x2 – y2 = (x + y) (x – y), we can write the given expression as;

(x + 3) (x – 3) = x2 – 32 = x2 – 9.

Problem: Solve (x + 5)3 using algebraic identities.

Solution: We know,

(x + y)3 = x3 + y3 + 3xy(x+y)

Therefore,

(x + 5)3 = x3 + 53 + 3.x.5(x+5)

= x3 + 125 + 15x(x+5)

= x3 + 125 + 15x2 + 75

= x3 + 15x2 + 200 (Answer)

Also Check:

Frequently Asked Questions – FAQs

What are Algebraic Identities?

Algebraic identities are the expressions or the equations of algebra which are true for every value of the variables present in the equation.

What are the three algebraic identities?

The three algebraic identities in Maths are:
Identity 1: (a+b)2 = a2 + b2 + 2ab.
Identity 2: (a-b)2 = a2 + b2 – 2ab.
Identity 3: a2 – b2 = (a+b) (a-b)

What is the value of (x + a) (x + b)?

Algebraic identities for (x + a) (x + b) = x2 + (a + b) x + ab.

What is the use of algebraic identities?

Algebraic Identities help in simplification. Algebraic Identities play a major role in solving quadratic equations and higher power equations.
The algebraic identities are also used to calculate products of numbers and other things.

How many total algebraic identities are there?

The standard algebraic identities are:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
a2 – b2 = (a + b)(a – b)
(x + a)(x + b) = x2 + (a + b) x + ab
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

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1 Comment

  1. please provide verifications of all identities, it will be helpful

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