Algebraic Identities

# Algebraic Identities Notes | Study Mathematics (Maths) Class 9 - Class 9

## Document Description: Algebraic Identities for Class 9 2022 is part of Mathematics (Maths) Class 9 preparation. The notes and questions for Algebraic Identities have been prepared according to the Class 9 exam syllabus. Information about Algebraic Identities covers topics like and Algebraic Identities Example, for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Algebraic Identities.

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Algebraic identities are an important set of formulas in maths. They form the foundation working principle of algebra and are helpful to perform computations in simple and easy steps. Certain algebraic problems require working across numerous mathematical step to obtain the answer. Here, with the use of algebraic identities, we are able to perform the calculations without any additional steps. Many of the algebraic identities have been obtained from the binomial expansion of terms.

An algebra identity means that the left-hand side of the equation is identically equal to the right-hand side, for all values of the variables. Here we shall try to acquaint ourselves with all the algebraic identities, their proofs, and how to use these identities in our math calculations.

What Are Algebraic Identities?

Algebraic identities are equations where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation. Unlike algebraic expressions, algebraic identities satisfy all the values of the variables. Let us consider an example to understand this better. Consider the equations: 5x - 3 = 12,  10x - 6 = 24, and x2 + 5x + 6 = 0. These equations satisfy only a unit value and do not work for any other values of the variables. Now let us consider an equation x2  - 9 = (x + 3)(x - 3). These equations satisfy any values of x.

From this, we can clearly understand that an algebra identity has an expression that satisfies any values for the variables. The basic algebraic identities are primarily helpful to work out the numerous math problems. The four basic algebraic identities are as follows.

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

(a - b)2 = a2 - 2ab + b2

(a + b)(a - b) = a2 - b2

(x + a)(x + b) = x2 + x(a + b) + ab

Algebraic Identities from Binomial Expansion Formula

The binomial expansion based on the binomial theorem is helpful to find derive all the algebraic identities. Let us gain a brief idea of the binomial theorem expansion, which would be helpful to derive algebraic identities of higher degrees. Generally, binomial expansion is used for expansion involving two variables and with n degree. The total number of terms of the binomial expansion is (n + 1) and in the expanded form the power of the variable is reducing, and the power of the second variable is increasing in sequential terms.  Each of the terms in the binomial expansion has a coefficient associated with it.

Binomial Expansion Formula

(x+y)n=nC0xny0+ nC1xn-1y1+ nC2xn-2y2 ...... nCn-1xyn-1 + nCnx0yn

The following algebraic identities have been derived from the binomial expansion for sum and difference of variable and for a maximum power of 3 the algebraic identities have been listed below as formulas. Further algebraic identities of higher degrees and more variables can also be derived using the above binomial expansion formula.

• (a + b)2 = a2 + 2ab + b2
• (a - b)2 = a2 - 2ab + b2
• (a + b)(a - b) = a2 - b2
• (a + b)3 = a3 +3a2b + 3ab2 + b3
• (a - b)3 = a3 - 3a2b + 3ab2 - b3
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

Algebraic Identities for Factorization

Algebraic identities are greatly helpful in easily factorizing algebraic expressions. The given expression representing the expanded form of the formula can be transformed and written as a set of factors using the below algebraic identities. Some of the higher algebraic identities such as a4 - b4 can be easily derived using the basic algebraic identities a2 - b2. The below list presents a set of algebraic identities helpful for the factorization of polynomials.

• a2 - b2 = (a - b)(a + b)
• x2 + x(a + b) + ab = (x + a)(x + b)
• a3 - b3 = (a - b)(a2 + ab + b2)
• a3 + b3 = (a + b)(a2 - ab + b2)
• a4 - b4 = (a - b)(a + b)(a2 + b2)
Algebraic Identities for Three Variables

The algebraic identities for three variables also has been derived using the binomial expansion formula. Further these identities are helpful to easily work across the algebraic expressions with the least number of steps.

• a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ac)
• a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ca - bc)
• (a + b)(a + c)(b + c) = (a + b + c)(ab + ac + bc) - 3abc

Apart from these simple algebraic identities listed above, there are other algebraic identities that we will use in higher grades.

Proof of Algebraic Identities

The following proofs of algebraic identities will help us to visually understand each of the identities and better understand it. Let us look at the proofs of each of the basic algebraic identities.

Proof of (x + a)(x + b) = x2 + x(a + b) + ab
(x+a)(x+b)  is nothing but the area of a rectangle whose sides are (x+a) and (x+b) respectively. The area of a rectangle with sides (x+a) and (x+b) in terms of the individual areas of the rectangles and the square is x2, ax, bx , ab.  Summing all these areas we have x2 + ax + bx + ab.  This gives us the proof for the algebra identity (x + a)(x + b) = x2 + ax + bx + ab = x2 + x(a + b) + ab.

Proof of (x+a)(x+b)

Proof of (a + b)2 = a2 + 2ab + b2

The algebraic expression (a+b)2 is nothing but (a+b) × (a+b). This can be visualized as a square whose sides are (a+b) and the area is (a+b)2. The square with a side of (a + b) can be visualized as four areas of a2, ab, ab, b2. The sum of these areas a2 + ab + ab + b2 gives the area of the square (a+b)2. The area of the square (a+b)2 =  a2 + ab + ab + b2 proves the algebraic identity.

Proof of (a + b)(a - b) = a2 - b2

The objective is to find the value a2 - b2, which can be taken as the difference of the area of two squares of sides a units and b units respectively.  This is equal to the sum of are areas of two rectangles as presented in the below figure. One rectangle has a length of a units and a breadth of (a - b) units. Another rectangle is taken with a length of (a - b) and a breadth of b units. Further, we take the areas of the two rectangles and sum the areas to obtain the resultant values. The respective areas of the two rectangles are (a - b) × a = a(a - b) , and (a - b) × b = b(a - b). Finally, we take the sum of these areas to obtain the resultant expression.

a(a + b) + b(a - b) = (a + b)(a - b)

Re-arranging the individual squares and rectangles, we get:

(a+b)(a−b)=a2−b2

Proof of (a − b)2 = a2 −2ab + b2

Once again, let’s think of (a - b)2 as the area of a square with length (a - b). To understand this, let's begin with a large square of area a2. We need to reduce the length of all sides by b, and it becomes a - b. We now have to remove the extra bits from a2 to be left with (a - b)2. In the figure below, (a - b)2 is shown by the blue area. To get the blue square from the larger orange square, we have to subtract the vertical and horizontal strips that have the area ab. However, removing ab twice will also remove the overlapping square at the bottom right corner twice. Hence, we add b2. Thus we have  (a − b)2 = a2 − ab − ab + b2. Hence this proves the algebraic identity (a − b)2 = a2 − 2ab + b2

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