GUNÌTA SAMUCCAYAH - SAMUCCAYA GUNÌTAH

In connection with factorization of quadratic expressions a sub-Sutra, viz. 'Gunita samuccayah-Samuccaya Gunitah' is useful.
It is intended for the purpose of verifying the correctness of obtained answers in multiplications, divisions and factorizations.

It means in this context: 'The product of the sum of the coefficients(sc) in the factors is equal to the sum of the coefficients(sc) in the product'

Symbolically we represent as sc of the product = product of the sc (in the factors)/p>


Example 1:     (x + 3) (x + 2) = x2 + 5x + 6
         Now ( x + 3 ) ( x + 2 ) = 4 x 3 = 12 : Thus verified.

Example 2:     (x + 5) (x + 7) (x - 2) = x3 + 10x2 + 11x - 70
         (1 + 5) (1 + 7) (1 - 2) = 1 + 10 + 11 - 70
         i.e.,     6 x 8 x -1 = 22 - 70
         i.e.,     -48 = -48 Verified.

Verify whether the following factorization of the expressions are correct or not by the Vedic check:
i.e. Gunita. Samuccayah-Samuccaya Gunitah:

   1.     (2x + 3) (x - 2) = 2x2 - x - 6
   2.     12x2 - 23xy + 10y2 = ( 3x - 2y ) ( 4x - 5y )
   3.     12x2 + 13x - 4 = ( 3x - 4 ) ( 4x + 1 )
   4.     ( x + 1 ) ( x + 2 ) ( x + 3 ) = x3 + 6x2 + 11x + 6

LOPANA STHÂPANÂBHYÂM

Lopana sthapanabhyam means 'by alternate elimination and retention'.

Consider the case of factorization of quadratic equation of type ax2 + by2 + cz2 + dxy + eyz + fzx. This is a homogeneous equation of second degree in three variables x, y, z. The sub-sutra removes the difficulty and makes the factorization simple. The steps are as follows:

i) Eliminate z by putting z = 0 and retain x and y and factorize thus obtained a quadratic in x and y by means of Adyamadyena sutra.

ii) Similarly eliminate y and retain x and z and factorize the quadratic in x and z.

iii) With these two sets of factors, fill in the gaps caused by the elimination process of z and y respectively. This gives actual factors of the expression.



Example 1 : 3x2 + 7xy + 2y2 + 11xz + 7yz + 6z2.


Step (i)  :   Eliminate z and retain x, y; factorize
             3x2 + 7xy + 2y2  = (3x + y) (x + 2y)
Step (ii) :   Eliminate y and retain x, z; factorize
             3x2 + 11xz + 6z2 = (3x + 2z) (x + 3z)
Step (iii):   Fill the gaps, the given expression
                              = (3x + y + 2z) (x + 2y + 3z)



Example 2 : 3x2 + 6y2 + 2z2 + 11xy + 7yz + 6xz + 19x + 22y + 13z + 20


Step (i)  :   Eliminate y and z, retain x and independent term
             i.e., y = 0, z = 0 in the expression (E).
             Then E = 3x2 + 19x + 20 = (x + 5) (3x + 4)
Step (ii) :   Eliminate z and x, retain y and independent term
             i.e., z = 0, x = 0 in the expression.
             Then E = 6y2 + 22y + 20 = (2y + 4) (3y + 5)
Step (iii):   Eliminate x and y, retain z and independent term
             i.e., x = 0, y = 0 in the expression.
             Then E = 2z2 + 13z + 20 = (z + 4) (2z + 5)
Step (iv) :   The expression has the factors
             (think of independent terms: constants)
                    = (3x + 2y + z + 4) (x + 3y + 2z + 5).


Solve the following expressions into factors by using appropriate sutras:

1. x2 + 2y2 + 3xy + 2xz + 3yz + z2.

2. 3x2 + y2 - 4xy - yz - 2z2 - zx.

ANTYAYOR DAŚAKE'PI

The Sutra simply means - numbers of which the last digits added up give 10.

This sutra is helpful in multiplying numbers whose last digits add up to 10(or powers of 10). The remaining digits of the numbers should be identical.

i.e. the Sutra works in multiplication of numbers for example:
25 and 25, 2 is common and 5 + 5 = 10
47 and 43, 4 is common and 7 + 3 = 10
62 and 68,
116 and 114.
and also for 425 and 475

The last example can be looked as digit 4 being common and 25 and 75 add upto 100.

Note that in each case the sum of the last digit of first number to the last digit of second number is 10. Further the portion of digits or numbers left wards to the last digits remain the same. At that instant use Ekadhikena on left hand side digits. Multiplication of the last digits gives the right hand part of the answer.
Example 1 : 57 X 53
See the end digits sum 7 + 3 = 10 ; then by the sutras Antyayor dasakepi and Ekadhikena we have the answer.
Ekadhikena to the remaining digits means, increment the remaining digits by 1 and multiply it with the same.
57 x 53 = ( 5 + 1 )x5 / 7x3 ( the '/' is just a seperator and not a division mark)&nbsp = 30 / 21
= 3021.

Example 2: 62 x 68
2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6.
Ekadhikena(increment) of 6 gives 7
62 x 68 = ( 6 x 7 ) / ( 2 x 8 )
= 42 / 16
= 4216.

Use Vedic sutras to find the products

1. 125 x 125
2. 34 x 36
3. 98 x 92

It is further interesting to note that the same rule works when the sum of the last 2, last 3, last 4 - - - digits added respectively equal to 100, 1000, 10000 -- - - . The simple point to remember is to multiply each product by 10, 100, 1000, - - as the case may be . Your can observe that this is more convenient while working with the product of 3 digit numbers.
Eg. 1: 292 x 208
Here 92 + 08 = 100, L.H.S portion is same i.e. 2
292 x 208 = ( 2 x 3 ) / 92 x 8
= 60 / 736 ( for 100 raise the L.H.S. product by 0 )
= 60736.

Find the following products using ‘Antyayordasakepi’

1. 318 x 312
2. 425 x 475

YAVADADHIKAM TAAVADAHIKIKRITYA VARGANCHA YOJAYET

This sutra means whatever the extent of its surplus, increment it still further to that very extent; and also set up the square of that surplus. This sutra is very useful in calculating the sqaures of numbers nearer(greater) to powers of 10.

For instance: in computing the square of 103 we go through the following steps:
1. The nearest power of 10 to 103 is 100.
2. Therefore, let us take 100 as our base.
3. Since 103 is 3 more than 100(base), we call 3 as the surplus.
4. Increase the given number further by an amount equal to the surplus.
i.e., perform ( 103 + 3 ) = 106. This is the left side of our answer!!.
5. On the right hand side put the square of the surplus, that is square of 3 = 09.
6. Append the results from step 4 and 5 to get the result.
Hence the answer is 10609.

Note: while calculating step 5, the number of digits in the squared number (09) should be equal to number of zeroes in the base(100). Hence in our case, the base 100 has 2 zeros and hence sqaure of 3 is 09 and not just 9.

Similarly if we need to calculate the square of 1009,
1. The nearest power of 10 to 1009 is 1000.
2. Therefore, let us take 1000 as our base.
3. Since 1009 is 9 more than 1000(base), we call 9 as the surplus.
4. Increase the given number further by an amount equal to the surplus.
i.e., perform ( 1009 + 9 ) = 1018. This is the left side of our answer!!.
5. On the right hand side put the square of the surplus, that is square of 9 = 081.(not just 81 !)
6. Append the results from step 4 and 5 to get the result.
Hence the answer is 1018081.

Try to find the square of 106 and 19 using this sutra.

Now here's a twist. What is sqaure of 112 using this rule? will this rule apply?
Answer is Yes. But we need to stick to basics and use some common sense.
Here we go...
1. & 2. the base is 100.
3. Surplus = 12
4. Increase the given number by surplus i.e. ( 112 + 12 ) = 124
5. Square of surplus, 12 = 144 . Now what?

by rule we need to consider only 2 digits in step 5 ( as the base has 2 zeros )
6. We have to append 124 and 144. The '1' in 144 becomes 'carry' to 124.
Hence the result is (124 +1 ) appended with 44 i.e., 12544 is the result.

dont get confused. try this example to make yourself more confident.
Find the square of 111. Now try to find the square of 10101.

YAVDUNAM TAAVDUNIKRITYA VARGANCHA YOJAYET

This sutra means whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency. This sutra is very handy in calculating squares of numbers near(lesser) to powers of 10

For instance in computing the square of 98 we go through the following steps:
1. The nearest power of 10 to 98 is 100.
2. Therefore, let us take 100 as our base.
3. Since 98 is 2 less than 100, we call 2 as the deficiency.
4. Decrease the given number further by an amount equal to the deficiency.
i.e., perform ( 98 -2 ) = 96. This is the left side of our answer!!.
5. On the right hand side put the square of the deficiency, that is square of 2 = 04.
6. Append the results from step 4 and 5 to get the result. Hence the answer is 9604.

Note: while calculating step 5, the number of digits in the squared number (04) should be equal to number of zeroes in the base(100). Hence in our case, the base 100 has 2 zeros and hence sqaure of 2 is 04 and not just 4.

Now consider another example: Calculate the square of 94. ( following the steps given above...)
1. The neareset power of 10 to given number 94 is 100
2. Hence the base is 100
3. Deficiency = 6
4. Subtract the deficiency from the given number ( 94 - 6 ) = 88
5. Square of deficiency = 36 ( 2 digits)
6. Append the results from step 4 and 5 to get the result. i.e., 8836

Try out calculating squares of 96, 994, 9999995

SAŃKALANA – VYAVAKALANĀBHYAM

This Sutra means 'by addition and by subtraction'. It can be applied in solving a special type of simultaneous equations where the x - coefficients and the y - coefficients are found interchanged.

Example 1:

45x – 23y = 113
23x – 45y = 91

From Sankalana – vyavakalanabhyam

add them,

i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91
i.e., 68x – 68y = 204 i.e., x – y = 3

subtract one from other,

i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91
i.e., 22x + 22y = 22 i.e., x + y = 1

and repeat the same sutra, we get x = 2 and y = - 1

Example 2:

1955x – 476y = 2482
476x – 1955y = -4913

Oh ! what a problem ! And still
just add, 2431( x – y ) = - 2431 i.e., x – y = -1
subtract, 1479 ( x + y ) = 7395 i.e., x + y = 5
once again add, 2x = 4 x = 2
subtract - 2y = - 6 y = 3

Solve the following problems using Sankalana – Vyavakalanabhyam.

1. 3x + 2y = 18
2x + 3y = 17

2. 5x – 21y = 26
21x – 5y = 26

Adyamadyena-Antyamantyena

The Sutra ' Adyamadyena-Antyamantyena' means 'the first by the first and the last by the last'.

Suppose we are asked to find out the area of a rectangular card board whose length and breadth are respectively 6ft . 4 inches and 5 ft. 8 inches. Generally we continue the problem like this.


Area = Length X Breath

By Vedic principles we proceed in the way "the first by first and the last by last"

i.e. 6’ 4" can be treated as 6x + 4 and 5’ 8" as 5x + 8,

Where x= 1ft। = 12 in; x2 is sq. ft.
Now ( 6x + 4 )(5x + 8 )



It is interesting to know that a mathematically untrained and even uneducated carpenter simply works in this way by mental argumentation. It goes in his mind like this

6’ 4"

5’ 8"

First by first i.e. 6’ X 5’ = 30 sq. ft.

Last by last i.e. 4" X 8" = 32 sq. in.

Now cross wise 6 X 8 + 5 x 4 = 48 +20 = 68.

Adjust as many '12' s as possible towards left as 'units' i.e. 68 = 5 X 12 +8 , 5 twelve's as 5 square feet make the first 30+5 = 35 sq. ft ; 8 left becomes 8 x 12 square inches and go towards right i.e. 8 x 12 = 96 sq. in. towards right ives 96+32 = 128sq.in.

Thus he got area in some sort of 35 squints and another sort of 128 sq. units.
i.e. 35 sq. ft 128 sq. in



Another Example:


Now 12 + 2 = 14, 10 x 12 + 24 = 120 + 24 = 144

Thus 4′ 6″ x 3′ 4″ = 14 Sq. ft. 144 Sq. inches.

Since 144 sq. in = 12 X 12 = 1 sq. ft .

The answer is 15 sq. ft.
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