Standard Identities In Algebra
INTRODUCTION
One of the most significant aspects of Elementary Mathematics is algebra. It begins in the lower grades of secondary school and continues through senior secondary and into higher education. There are a thousand issues to discuss when it comes to the relevance of Algebraic Identities in Math.
- Definition
Algebraic identities are algebraic equations that are valid for all values of variables in them. Algebraic equations are mathematical expressions made up of integers, variables (unknown values), and operators (addition, subtraction, multiplication, division, etc.)
Algebraic identities are utilised in a variety of mathematical fields, including algebra, geometry, and trigonometry. These are mostly used to find the polynomial factors. The polynomial factorisation is solved using it.
Algebraic identities are utilised in this way to compute algebraic expressions and solve various polynomials.
On both sides of the polynomial, i.e. Left Hand Side and Right Hand Side, they include variable and constant. LHS must equal RHS in algebraic identity.
- What do you mean by identity?
To understand Algebraic Identity better, let us consider the linear equation ax+b=0.
We all know the main property of Algebraic Identity is LHS – RHS.
Here, the left-hand side and right-hand sides of the above equations are the same only.
When x = – ba Hence, it is not identity, but it is an equation.
In (a+b)2=a2+b2+2ab, we know that it is true for all values of a and b. So, it is an identity.
Example
Consider the equality (a + 2) (a + 3) = a2 + 5a + 6.For any value of a, both sides of this equality can be evaluated.
For a = 6,
LHS = (a + 2) (a + 3)
= (6 + 2) (6 + 3)
= 8 × 9
= 72
RHS = a2 + 5a + 6
= ( 6 )2 + 5 × 6 + 6
= 36 + 30 + 6
= 72
As a result, when a= 6, the values of the two sides of the equality are equal. LHS = RHS may be found for any value of a. An identity holds for all possible values of a given variable. As a result, (a + 2) (a + 3) = a2 + 5a + 6
Thus we can say ( a+2)( a+ 3)= a2 +5a +6 is an identity.
List of Standard Algebraic Identities List
The Binomial Theorem is the source of common algebraic identities. Expanding the powers of binomials or the total of two terms yields the binomial theorem, often known as binomial expansion. Binomial coefficients are the coefficients used in conjunction with the terms for expanding.
Now the question arises how many algebraic identities are there.
Following is a list of a few standard algebraic identities
1st Identity (Square of Sum of Two Terms )
(a + b)2 = a2 + 2ab + b2 |
2 nd Identity (Square of Difference of Two Terms )
(a – b)2 = a2 – 2ab + b2 |
3 rd Identity (Difference of Two Squares)
a 2– b 2= (a +b) (a-b) |
4 th Identity
(x + a)(x + b)=x2+(a+b)x +ab |
5thIdentity (Cube of Difference of Two Terms)
(a – b )3 = a3 + b3 + 3ab (a +b) |
6 th Identity (Square of the sum of three Terms)
(a +b + c )2 = a2 + b2 +c 2+2ab +2bc +2ca) |
7 th Identity ( (Cube of sum of Two Terms)
(a + b)3 = a3 + b3 + 3ab (a + b) |
8th Identity
(a – b)3 = a3 – b3 – 3ab (a – b) |
9 th Identity(involves Trinomial)
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) |
10 th Identity( Difference of cube of two terms)
a3 – b3 =(a – b)(a2 +ab + b2) |
11 th Identity( Sum of cube of two terms)
a3 + b3 =(a + b)(a2 +ab + b2) |
12 th Identity
a4 – b4 =( a2 + b2)( a2 – b2) |
SOLVING ALGEBRAIC IDENTITIES
The easiest way to solve Algebraic Identities is the
- Substitution Method
Substitution Method: The substitution approach, in which we substitute values in variable places and strive to make both sides equal, may be used to check algebraic identities. LHS Equals RHS, for example.
Example 1
(x – 5) (x + 5) = x2 – 52
We’ll begin by substituting value for x.
starting with x = 1, (-4) x (6) = -24
Now will repeat the same on the RHS
Put x = 1 , 12 – 52 = -24
Similarly put x = 2 , 22 -52 = – 21
then we will put x = 2, (-3) x 7 = – 21
Here we got x = 1 and x = 2 as the value which satisfy the given question.
Example 2
(x + 4) (x + 5) = x2 + 9x +20
We’ll begin by substituting value for x on LHS
starting with x = 1, ( 5 ) x ( 6 ) = 30
then we will put x = 2, ( 6 ) x 7 = 42
Now will repeat the same on the RHS
Put x = 1 , 12 +9(1) +20 = 30
Similarly put x = 2 ,22 +9(2) +20 = 4 + 18 +20 = 42
Here we got x = 1 and x = 2 as the value which satisfy the given question.
SOLVED EXAMPLES FOR ALGEBRAIC IDENTITY
Example 1: Simplify ( 3x +5y )2 + ( 3x – 5y )2
Solution:The following algebraic identities must be used to solve this:
(a + b)2 = a2 + 2ab + b2…………….(I)
(a – b)2 = a2 – 2ab + b2………………(II)
Adding (I) and (II) we have:
(a + b)2 + (a – b)2 = a2 + 2ab + b2 + a2 – 2ab + b2
(a + b)2 + (a – b)2 = 2a2 +2 b2…………….(III)
Here we have a = 3x and b = 5y.
Substituting the value of a and b in the expression( III) we have:
( 3x +5y )2 + ( 3x – 5y )2 = 2(3x)2 + 2(5y)2
= 18x2 + 50y2
Answer: (3x + 5y)2 + (3x – 5y)2 = 18x2 + 50y2
Example 2: The area of a square is 9x2 +12x +4. What is the measure of the side of the square?
Solution:
This looks like the RHS of the algebraic identity: (a + b)2 = a2 + 2ab + b2
Comparing this with the given expression 9x2 +12x +4,
we have a2 = 9x2 so a = 3x,
b2 = 4 so b = 2
9x2 +12x +4 = (3x)2 + 2(3x)(2) + 22
= (3x+2)2
Therefore, the Area of the square = product of its sides=(3x +2) x (3x +2)
Answer: Therefore, the side of the square is (3x + 2).
Example 3: Using identities, solve 197 × 203.
Solution: 197 × 203 can also be written as ( 200 – 3 ) × ( 200 + 3 )
This is based on the algebraic identity (a + b)(a – b) = a2 – b2………..(I)
Here we have a = 200, and b = 3
Substituting the values in ( I ), we get:
(200 – 3)(200 + 3) = ( 200 )2 – ( 3 )2
= 40000 – 9
= 39991
Answer: Therefore 197 × 203 = 39991
Example 4 : Using conventional algebraic identities, Factorise 25×2+16y2+9z2–40xy+24yz–30zx
Solution:
Given expression :25x2 +16y2 + 9z2 –40xy + 24yz – 30zx
⇒(5x)2 +( 4y)2 +( 3z)2 – 40xy + 24yz – 30zx
⇒(–5x)2+(4y)2+(3z)2 + 2×(–5x)×4y + 2×4y×3z + 2×(–5x)×3z……………(I)
Using the identity: (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca……………(II)
Comparing equation (I) and (II) we have a = –5x and b = 4y
and c=3z
⇒(–5x+4y+3z)2
Hence, the factors of 25×2+16y2+9z2–40xy+24yz–30zx
are (4y+3z–5x)(4y+3z–5x).
Example 5 : Expand ( 5 x – 6y) 3 using a standard algebraic identity.
Solution
The above equation can be solved using the identity
(a – b)3 = a3 – b3 – 3ab (a – b)……………….(I)
Now a = 5x and b = 6y
Putting values of a and b in the equation (I)
= (5x)3 – (6y)3 – 3(5x)(6y) [ 5x -6y]
= 125 x3 – 216 y 3 – 90 xy[ 5x -6y]
= 125 x3 – 216 y3 -450x2y + 540 xy2
Answer: Hence the expansion of (5x -6y)3 is 125 x3 – 216 y3 -450x2y + 540 xy2
CONCLUSION
We looked at what algebraic identities are and how crucial they are in solving problems in mathematics. We also spoke about several common identities and how to prove them. Furthermore, these identities serve as the foundation for all algebraic formulae. Finally, we went through several instances that may help you comprehend these identities.