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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 = ½ But by Vilokanam i.e.,, observation 1 5 x + __ = __ can be viewed as x 2 1 1 x + __ = 2 + __ giving x = 2 or ½. 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 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 |
