Algebraic Identities
Algebraic Identity Definition:
An Identity is an equality which is true for every value of the variable
in it.
Example: ( x+1 ) ( x+2
) = x2 + 2x + x + 2
=
x2 + 3x + 2
For any value of x LHS is equal to RHS,which shows the appearance of identity
here.
Some of the identities helpful for solving the problems are given below:
- (a
+ b )2 = a2 + 2ab + b2
- (a
- b )2 = a2 -2ab + b2
- (a
+ b ) ( a - b ) = a2 - b2
Proof of Identity
( a + b)2
Step 1: expand the term
= ( a + b) ( a +
b)
Step 2: factories
= a ( a + b) + b ( a + b)
Step 3: simplify = a2 +
ab + ba + b2
Step 4: add the common term = a2 + 2ab + b2
assign a = 1, b = 2
a + b)2 = a2 + 2ab + b2
(1 + 2)2 =
12 + 2*1*2 + 22
(3)2
= 1 + 4 + 4
9 = 9
LHS = RHS
List of algebraic identity :
The following are some of the important algebraic identities or
expression used in class 9th maths
1. (a + b)2 = a2 + 2ab + b2
2. ( a - b)2 = a2 - 2ab + b2
3. (a + b ) ( a - b ) = a2 - b2
4. ( x + a ) ( x + b ) = x2 + ( a + b) x + ab
5. (x + a ) ( x - b) = x2 + ( a -b )
x - ab
6. ( x -a ) ( x + b ) = x2 + ( b - a )
x - ab
7. ( x - a ) ( x - b ) = x2 - ( a + b
) x + ab
8. ( a + b )3 = a3 + b3
+ 3ab ( a + b )
9. ( a - b )3 = a3 - b3 -
3ab (a - b )
10. (x + y + z)2 = x2 + y2 +
z2 + 2xy +2yz + 2xz
11. (x + y - z)2 = x2 + y2 +
z2 + 2xy - 2yz - 2xz
12. ( x - y + z)2 = x2 + y2 +
z2 - 2xy - 2yz + 2xz
13. (x - y - z)2 = x2 + y2 +
z2 - 2xy + 2yz - 2xz
14. x3 + y3 + z3 -
3xyz = (x + y + z ) ( x2 + y2 + z2 -
xy - yz -xz)
15. x2 + y2 = 12 [( x
+ y)2 + ( x - y)2]
16. ( x + a) ( x + b) ( x + c) = x3 +
(a + b + c) x2 + ( ab + bc + ca ) x
+ abc
17. x3 + y3 = (x + y) ( x2 -xy
+ y2 )
18. x3 - y3 = ( x -
y) ( x2 + xy + y2 )
19. x2 + y2 + z2 -xy
- yz - zx = 12 [( x - y)2 + (y -z)2 +
( z - x)2]
Examples based on Identity:
Example 1: Factorize the term (xy)2 –
82
Solution: Given (xy)2 –
82
Step 1: First check and make use of
identity (a + b) (a - b) = a2 -b2
Step 2: Assign the value in
identity
( xy )2 – 82
= ( xy + 8 ) ( xy - 8 )
Example 2: Expand ( x - 2y)2
Solution :
Step 1: Make use of the identity ( a -b )2 =
a2 - 2ab + b2
Step 2: Assign the value in equation
a = x, b = 2y = ( x)2 - 2 * x * 2y +
(2y)2
Step 3: Simplify =
x2 - 4xy + 4y2
Important
Concepts and Tips to Solve Quadratic Equations
Structure of a quadratic
equation = X²± (Sum of Root) X ± (Product of root) = 0
In the question discussed
below the coefficient of X²≠ 1
To solve these types of
questions, PR (Product of root) will be taken as (PR × coefficient of X²)
And X = X value /
coefficient of X²
DIRECTIONS
In each question below one
or more equations are given on the basis of which we are supposed to find out
the relationship between x and y
Give answer (1) if X>Y
Give answer (2) if X≥Y
Give answer (3) if X<Y
Give answer (4) if X≤Y
Give answer (5) if X=Y or
the relationship cannot be determined
QUESTION:
(i) 10X²– 7X + 1 = 0
(ii) 35Y²– 12Y + 1 = 0
GIVEN:
If the given SR is –ve
then consider it as +ve
If the given SR is +ve
then consider it as –ve
In equation (i)
Sum of Root (SR) = +7
Product of Root (PR) = 10
i.e., (1 × 10 = PR × co-efficient of X²)
Similarly in eq. (ii)
SR = +12
PR = 35 because (1 × 35 =
PR × co-efficient of Y²)
SOLUTION
(i) 10X²– 7X + 1 = 0
SR = +7
X = 5, 2
PR = 10, 5×2
Split the PR into its
divisible numbers such that when the numbers are added or subtracted we get the
SR
Here 5 × 2 = 10 (PR)
And 5 + 2 = 7 (SR)
In this type of quadratic
equation, where the coefficient of X²≠ 1
X = X value / coefficient
of X²
X = (5, 2) = ([5/10],
[2/10]) = (0.5, 0.2)
Therefore, X = 0.5, 0.2
(ii) 35Y²– 12Y + 1 = 0
SR = +12
Y = 7, 5
PR = 35
7×5
Here 7 × 5 = 35 (PR)
And 7 + 5 = 12 (SR)
Here the coefficient of Y²≠
1
Y = Y value / coefficient
of Y²
Y = (7, 5) = ([7/35],
[5/35]) = (0.2, 0.14[approx.])
Therefore, Y = 0.2, 0.14
We have calculated the
values of X and Y, now we have to compare the values with each other to decide the
relation between them
X = 0.5, 0.2; Y = 0.2,
0.14
Take X = 0.5, compare it
with both the values of Y = 0.2, 0.14
We get, X = 0.5 is greater
than Y = 0.2 i.e., X>Y
X = 0.5 is greater than Y
= 0.14 i.e., X>Y
Similarly Take X = 0.2,
compare it with both the values of Y = 0.2, 0.14
We get, X = 0.2 is equal
to Y = 0.2 i.e., X=Y
X = 0.2 is greater than Y
= 0.14 i.e., X>Y
So the relation between X
and Y is given by both X = Y and X>Y i.e., X≥Y
Therefore Answer is (2) if
X≥Y
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