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 rule 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 the variables.
Algebraic Identities for Ttwo 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.
- (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.
- (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.
- 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
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.
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)
(x + 5)3 = x3 + 53 + 3.x.5(x+5)
= x3 + 125 + 15x(x+5)
= x3 + 125 + 15x2 + 75
= x3 + 15x2 + 200 (Answer)
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