|
|
||
|
Identities - Part I |
||
|
|
||
|
In
algebraic equations one can find the values of variables, if the equalities
are valid. In case of more than one variable, we can solve simultaneous
equations and the values of the variables for which the given equations are
correct. There
are a few algebraic equations, which remain valid for all values of
variables. Such equations are called identities.
We will study a few of these identities in this chapter. What
we will study in this chapter
: 1.
Identity for the square of sum of two terms 1. Identity for the square of sum of two terms If
a and b are two variables or terms, then the identity for the sum of a and b
is (a
+ b)2 = a2+ 2ab + b2 Let
us try and prove this identity. (a
+ b)2 = (a + b) * (a
+ b)
= a (a + b) + b(a + b)
= a2 + ab + ba
+ b2 Since
ab = ba
To
if this is an identity let
us put a few values of a and b and see if the identity is true. Example 1:
Let a = 2, b = 1. Prove that (a
+ b)2 = a2 + 2ab
+ b2 LHS
: ( 2 + 1)2 = (3) 2 =
9 RHS
: 22 + 2 * 2 * 1 + 12
= 4 + 4+ 1 = 9 Thus
LHS = RHS The
equation (a + b)2 =
a2 + 2ab
+ b2 is valid for a = 2 and b = 1 Example 2
: Let a = 10, b = 5. Prove that
(a + b)2 = a2 + 2ab
+ b2 LHS
: ( 10 + 5)2 = (15)
2 = 225 RHS
: 102 + 2 * 10 * 5 + 52 = 100 + 100+ 25 = 225 Thus
LHS = RHS The
equation (a + b)2 =
a2 + 2ab
+ b2 is valid for a = 10 and b = 5 Example 3
: Let a = 0.5, b = 1. Prove
that (a + b)2 = a2 + 2ab
+ b2 LHS
: ( 0.5 + 1)2 =
(1.5) 2 = 2.25 RHS
: 0.52 + 2 * 0.5 * 1 + 12 = 0.25 + 1 + 1 = 2.25 Thus
LHS = RHS The
equation (a + b)2 =
a2 + 2ab
+ b2 is valid for a = 0.5 and b = 1 From
the above three examples, we can see that the identity (a + b)2
= a2 +
2ab + b2
remains valid for any values of a and b including fractions or
decimals numbers. Thus,
the square of the sum of two terms = the square of the first term +
twice the product of the two terms + the square of the second term. The
identity can be proven by use of geometry also : Consider a square ABCD
whose side is (a +b).
Area
of
The
The area
+
area of rectangle HPGD +
area of the square PFCG = area I + area II + area III + area IV area
I = a * a = a2 area
II = a * b = ab area
III = a * b = ab area
IV = b2 Thus
the area of square ABCD = (a + b)2
= a2 +
2ab + b2 Example
4
: Expand (4x + y)2 Using
the identity (a + b)2 =
a2 + 2ab +
b2 we can write (4x
+ y)2 = (4x)2
+ 2 * 4x * y
+ y2
= 16x2 + 8xy
+ y2 Thus
(4x + y)2 = 16x2
+ 8xy + y2 Example
5
: Find the square of 42. 422
= (40 + 2)2
= 402 + 2 * 40 * 2 + 22
= 1600 + 160 + 4
= 1764 Thus
422 = 1764 Example
6
: Find the square of 210. 2102
= (200 + 10)2
= 2002 + 2 * 200 * 10 + 102
= 40000 + 4000 + 100
= 44100 Thus
2102 = 44100 In
examples 5 and 6 it is shown how squares
of large numbers can be calculated easily by writing the number as sum of
two easy numbers.
|
||
|
||
|
|
