Methods to Solve Algebraic Identities
- We can verify algebraic identities by substitution method, in which we can put values in variable places and try to make both sides equal. i.e LHS = RHS.
Example:
(a – 2) (a + 2) = a2 – 22
Now we will start putting value in place of a.
starting with a = 1, (-1) x (3) = -3
then we will put a = 2, 0 x 4 = 0
Here we got a = 1 and a = 2 as the value which satisfy the given question.
- Another method is by manipulating identities which are commonly used:
i. (a + b)2 = a2 + b2 + 2ab
ii. (a – b)2 = a2+ b2 – 2ab
iii. (a + b)(a – b) =a2 – b2
iv. (x + a)(x + b) = x2 + (a + b)x + ab
Proof:
i. (a + b)2 = (a + b) (a + b)
= (a + b) (a) + (a + b) (b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
Hence, LHS = RHS.
ii. (a – b)2 = (a – b) (a – b)
= (a – b) (a) + (a – b) (b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Hence, LHS = RHS.
iii. (a + b) (a – b) = a (a – b) + b (a – b)
= a2 – ab + ab – b2
= a2 – b2
Hence, LHS = RHS.
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Standard Algebraic Identities
Algebraic Identities are algebraic equations that are always true for every value of the variable in them. The algebraic equations that are valid for all values of variables in them are called algebraic identities.
It is used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and in solving different polynomials. They contain variable and constant on both the side of polynomial i.e. LHS and RHS. In algebraic identity, LHS must be equal to RHS.
This article provides you with the standard algebraic identities, including their examples, and methods to solve algebraic identities.
Table of Content
- What are Identities?
- Standard Algebraic Identities List
- Methods to Solve Algebraic Identities
- Standard Algebraic Identities Examples
- Standard Algebraic Identities Practice Problems