maths

VILOKANAM

The Sutra 'Vilokanam' means 'Observation'. Generally we come across problems which can be solved by mere observation. But we follow the same conventional procedure and obtain the solution. But the hint behind the Sutra enables us to observe the problem completely and find the pattern and finally solve the problem by just observation.

Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the problem, the conventional process tends us to solve the problem in the following way.

1 5
x + __ = __
x 2

x2 + 1 5
_____ = __
x 2

2x2 + 2 = 5x
2x2 – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
2x (x – 2) – (x – 2) = 0
(x – 2) (2x – 1) = 0
x – 2 = 0 gives x = 2
2x – 1 = 0 gives x = Ѕ

But by Vilokanam i.e.,, observation

1 5
x + __ = __ can be viewed as
x 2

1 1
x + __ = 2 + __ giving x = 2 or Ѕ.
x 2

Consider some examples.

Example 1 :
x x + 2 34
____ + _____ = ___
x + 2 x 15

In the conventional process, we have to take L.C.M, cross-multiplication. simplification and factorization. But Vilokanam gives

34 9 + 25 3 5
__ = _____ = __ + __
15 5 x 3 5 3

x x + 2 3 5
____ + _____ = __ + __
x + 2 x 5 3

gives
x 3 5
_____ = __ or __
x + 2 5 3

5x = 3x + 6 or 3x = 5x + 10
2x = 6 or -2x = 10
x = 3 or x = -5


Example 2 :

x + 5 x + 6 113
____ + _____ = ___
x + 6 x + 5 56

Now,
113 49 + 64 7 8
___ = _______ = ___ + ___
56 7 x 8 8 7

x + 5 7 x+5 8
____ = __ or ____ = __
x + 6 8 x+6 7

8x + 40 = 7x + 42 7x + 35 = 8x + 48
or
x = 42 - 40 = 2 -x = 48 – 35 = 13
x = 2 or x = -13.

Example 3:

5x + 9 5x – 9 82
_____ + _____ = 2 ___
5x - 9 5x + 9 319

At first sight it seems to a difficult problem.

But careful observation gives

82 720 841 - 121 29 11
2 ___ = ___ = ________ = ___ - __
319 319 11 x 29 11 29

(Note: 292 = 841, 112 = 121)

5x + 9 29 -11
_____ = __ or ___
5x - 9 11 29

(Note: 29 = 20 + 9 = 5 x 4 + 9 ; 11 = 20 – 9 = 5 x 4 – 9 )

i.e.,
x = 4 or
5x + 9 -11
_____ = ___
5x - 9 29

145x + 261 = -55x + 99
145x + 55x = 99 – 261
200x = -162

-162 -81
x = ____ = ____
200 100

Simultaneous Quadratic Equations:

Example 1: x + y = 9 and xy = 14.

We follow in the conventional way that

(x – y)2 = (x + y)2 – 4xy = 92 – 4 (14) = 81 - 56 = 25
x – y = √ 25 = ± 5



x + y = 9 gives 7 + y = 9
y = 9 – 7 = 2.

Thus the solution is x = 7, y = 2 or x = 2, y = 7.

But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. Hence the solution.

Example 2: 5x – y = 7 and xy = 6.

xy = 6 gives x = 6, y = 1; x = 1, y = 6;

x = 2, y = 3; x = 3, y = 2 and of course negatives of all these.

Observe that x = 6, y = 1; x = 1, y = 6: are not solutions because they do not satisfy the equation 5x – y = 7.

But for x = 2, y = 3; 5x – y = 5 (2) – 3 = 10 – 3 = 7 we have 5(3)–2≠7.

Hence x = 2, y = 3 is a solution.

For x = 3, y = 2 we get 5 (3) – 2 = 15 – 2 ≠ 7.

Hence it is not a solution.

Negative values of the above are also not the solutions. Thus one set of the solutions i.e., x = 2, y = 3 can be found. Of course the other will be obtained from solving 5x – y = 7 and 5x + y = -13.

i.e., x = -3 / 5, y = -10.

Partial Fractions:

Example 1: Resolve

2x + 7
___________ into partial fractions.
(x + 3) (x + 4)

2x + 7 A B
We write ____________ = ______ + ______
(x + 3)(x + 4) (x + 3) (x + 4)

A (x + 4) + B (x + 3)
= __________________
(x + 3) (x + 4)

2x + 7 ≡ A (x + 4) + B (x + 3).

We proceed by comparing coefficients on either side

coefficient of x : A + B = 2 ..........(i) X 3

Independent of x : 4A + 3B = 7 .............(ii)

Solving (ii) – (i) x 3 4A + 3B = 7
3A + 3B = 6
___________
A = 1

A = 1 in (i) gives, 1 + B = 2 i.e., B = 1

Or we proceed as

2x + 7 ≡ A (x + 4) + B (x + 3).
Put x = -3, 2 (-3) + 7 ≡ A (-3 + 4) + B (-3 + 3)
1 = A (1) ... A = 1.

x = -4, 2 (- 4) + 7 = A (-4 + 4) + B (-4 + 3)
-1 = B(-1) ... B = 1.

2x + 7 1 1
Thus ____________ = _____ + _____
(x + 3) (x + 4) (x + 3) (x + 4)

2x + 7
But by Vilokanam ____________ can be resolved as
(x + 3) (x + 4)

(x + 3) + (x + 4) =2x + 7, directly we write the answer.

Example 2:

3x + 13
____________
(x + 1) (x + 2)

from (x + 1),(x + 2) we can observe that

10 (x + 2) – 7(x + 1) = 10x + 20 – 7x – 7 = 3x + 13

3x + 13 10 7
Thus ____________ = _____ - _____
(x + 1) (x + 2) x + 1 x + 2

Example 3:

9
________
x2 + x - 2

As x2 + x – 2 = (x – 1) (x + 2) and
9 = 3 (x + 2) – 3 (x – 1)
(3x + 6 – 3x + 3 = 9)


9 3 3
We get by Vilokanam, ____________ = ____ - ____
x2 + x – 2 x - 1 x + 2


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I. Solve the following by mere observation i.e. vilokanam

1. 2.
1 25 1 5
x + __ = __ x - __ = __
x 12 x 6

3.
x x + 1 1
_____ + _____ = 9 __
x + 1 x 9

4.
x + 7 x + 9 32
____ - ____ = ___
x + 9 x + 7 63

II. Solve the following simultaneous equations by vilokanam.

1. x – y = 1, xy = 6 2. x + y = 7, xy = 10

3. 2x + 3y = 19, xy = 15

4. x + y = 4, x2 + xy + 4x = 24.

III. Resolve the following into partial fractions.

1.
2x - 5
____________
(x – 2) (x – 3)

2.
9
____________
(x + 1) (x – 2)

3.
x – 13
__________
x2 - 2x - 15

4.
3x + 4
__________
3x2 + 3x + 2