You can find all my previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics

A Spectacular Illustration of Vedic Mathematics

10's Complements

Multiplication Part 1

Multiplication Part 2

Multiplication Part 3

Multiplication Part 4

Multiplication Part 5

Multiplication Special Case 1

Multiplication Special Case 2

Multiplication Special Case 3

Vertically And Crosswise I

Vertically And Crosswise II

Squaring, Cubing, Etc.

Subtraction

Division By The Nikhilam Method I

Division By The Nikhilam Method II

Division By The Nikhilam Method III

Division By The Paravartya Method

Digital Roots

Straight Division I

Straight Division II

Vinculums

Divisibility Rules

Simple Osculation

Multiplex Osculation

Solving Equations 1

Solving Equations 2

Solving Equations 3

Solving Equations 4

Mergers 1

Mergers 2

Mergers 3

Multiple Mergers

Complex Mergers

Simultaneous Equations 1

Simultaneous Equations 2

Quadratic Equations 1

Quadratic Equations 2

Quadratic Equations 3

The first type of non-standard quadratic equation was already dealt with in this earlier lesson in the context of the Sunyam Samyasamuccaye sutra. These are equations in which the sum of the terms in the numerator is the same as the sum of the terms in the denominator, and the difference between numerator and denominator of each term is the same on both sides of the equal to sign. An example of such an equation is:

(2x + 1)/(3x + 2) = (5x + 2)/(4x + 1)

You will immediately notice that the sum of the denominators of both terms on either side of the equal to sign is 7x + 3. Similarly, the sum of the numerators of both terms on either side of the equal to sign is also 7x + 3. The Sunyam Samyasamuccaye sutra says that in that case, this sum can be set to zero for one solution of the equation. Thus, 7x + 3 = 0, giving us x = -7/3 as one solution. We also see that the difference between the numerator and denominator on either side of the equal to sign is x + 1. Application of the sutra then tells us that this difference can be set to zero for the other solution to the equation. Thus, x + 1 = 0, giving us x = -1 as the other solution.

As we saw in the earlier lesson, this is very useful in solving equations that are expressed as below:

(ax + b)^2 = (cx + d)^2

where a is not equal to c.

We can then rewrite the equation mentally, as below:

(ax + b)/(cx + d) = (cx + d)/(ax + b)

We know that in this form, the sunyam samyasamuccaye sutra is applicable in both its meanings, giving us two solutions of the quadratic equation immediately. One meaning tells us that (a + c)x + b + d = 0, which gives us x = -(b + d)/(a + c). The other meaning tells us that (a - c)x + b - d = 0, giving us x = (d - b)/(a - c).

For instance, consider the equation (3x + 5)^2 = (2x + 1)^2. We can spend a lot of time expanding out the terms, collecting them on one side and deriving an equation that is expressed in standard form, and then apply the quadratic formula to it. Or, we could recognize that this is a case where the sunyam samyasamuccaye sutra is applicable, and derive the two roots mentally as x = -6/5 and x = -4!

Now consider another equation of the following type:

a/(x + a) + b/(x + b) = c/(x + c) + d/(x + d)

Also assume that a + b = c + d. In that case, the sunyam samyasamuccaye sutra tells us that one of the roots is x = 0, and the other root is derived by setting (x + a) + (x + b) = 0. Why? Consider the following manipulation of the first term, a/(x + a).

a/(x + a) = 1 - x/(x + a)

By similar manipulation of all the terms, we can rewrite the equation as:

1 - (x)/(x + a) + 1 - (x)/(x + b) = 1 - (x)/(x + c) + 1 - (x)/(x + d)

Cancelling out the 1's from both sides of the equation, we get:

x/(x + a) + x/(x + b) = x/(x + c) + x/(x + d)

This then tells us that x = 0 because it is a common factor in all terms of the equation (from this previous lesson). We can then remove the x from the numerator of the equation leaving us with:

1/(x + a) + 1/(x + b) = 1/(x + c) + 1/(x + d)

Since a + b = c + d, the sum of the denominators on the left-hand side of the equation is the same as the sum of the denominators on the right hand side of the equation. Equating this sum to zero gives us the other root (this result is from this previous lesson). Thus, setting 2x + a + b = 0 gives us x = -(a + b)/2 as the other root of the equation.

Consider the equation below:

1/(x + 1) + 4/(x + 4) = 2/(x + 2) + 3/(x + 3)

We find that this follows the pattern we have dealt with just now, and also 1 + 4 = 2 + 3. We can then say, on sight, that the solutions to this equation are x = 0, and x = -2.5.

The next special type of equation we will consider is very similar to the merger type of equation we dealt with in this previous lesson. However, the sum of the numerators on the left hand side of the equation is not equal to the numerator on the right hand side, as required for the merger operation. Instead, consider the equation below:

a/(x + b) + c/(x + d) = e/(x + f)

where a/b + c/d = e/f.

When the above condition is satisfied, then one of the roots is zero, and the other root can be derived by merger. Let us see how.

a/(x + b) + c/(x + d) = e/(x + f) can be manipulated to

a/b - (ax/b)/(x + b) + c/d - (cx/d)/(x + d) = e/c - (ex/f)/(x + f)

Since a/b + c/d = e/f, we can cancel those terms out of the equation above, giving us:

(ax/b)(x + b) + (cx/d)/(x + d) = (ex/f)/(x + f)

Now, since x is a common term in all three numerators, one of the roots is x = 0. Removing the x from the numerator of all three terms gives us:

(a/b)/(x + b) + (c/d)/(x + d) = (e/f)/(x + f)

Once again, since a/b + c/d = e/f, the above equation can be solved by merger operation easily. Consider the example equation below:

6/(x + 3) + 4/(x + 2) = 20/(x + 5)

We apply the test above, and find that 6/3 + 4/2 = 20/5. Therefore, we can set one root to x = 0, and then simplify the equation as below:

2/(x + 3) + 2/(x + 2) = 4/(x + 5)

The new numerators are a/b, c/d and e/f as worked out in the algebraic equation. We can then solve this by merger by setting p = 2, q = 2, a = 3, b = 2 and c = 5. The solution is x = -13/5. Thus the two roots of the given equation are x = 0 and x = -13/5.

Before I leave, I want to touch upon a set of equations that may be presented in standard form, but which may be easier to solve by converting to a special type. These are equations of the type ax^2 + bx + c = 0 where a and c are both 1. Thus the equations have the form x^2 + bx + 1 = 0. It is easy to see that these equations can easily be converted to the following form:

x + 1/x = -b

As we saw in the previous lesson, this kind of equation is easy to derive approximate roots for mentally. For instance, if you are faced with the equation x^2 -10x + 1 = 0, you can instantly estimate the roots to be x = 9.90 and x = 0.10. The exact roots turn out to be 0.1010 and 9.8990!

Similarly, if the equation in standard form is presented as x^2 + bx - 1 = 0, then it is easy to see that this can be converted to x - 1/x = -b. This, again, can be solved mentally as we saw in the previous lesson to derive approximate roots. Thus, if asked to solve x^2 + 10x - 1 = 0, we can quickly say that the roots are approximately 0.10 and -10.10. The exact roots turn out to be 0.0990 and -10.099!

Hope this lesson (combined with the previous lessons on quadratic equations) has provided you with a little more insight into how certain quadratic equations can be solved without going through the trouble of expressing them in standard format and applying the quadratic formula. We have also dealt with a couple of special cases of equations that are expressed in standard form being more amenable to mental solution than application of the quadratic formula. Being able to recognize the applicability of these techniques is very important to the time and labor savings associated with these techniques. Please be sure to practice these techniques so that their application comes as naturally as the application of the quadratic formula! Good luck, and happy computing!!

## No comments:

Post a Comment