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This post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! I have also written another short story and posted it on my blog here. If you can provide me feedback or have other thoughts on my writing (content, style, anything at all), I would appreciate hearing from you. Thank you!

In the previous lesson, we learned about the solution technique for a type of quadratic equation which consists of a sum or difference of reciprocals on each side of the equal-to sign. We found that solving such equations can be much easier than actually expanding out the terms, collecting like terms together, brining the quadratic equation into the standard form, and then applying the quadratic formula. However, the main obstacle to applying this method to the solution of many such equations is that the right-hand side of the equation may not be readily capable of being split into a pair of reciprocals, similar to the left-hand side.

In this lesson, we will assume that the left-hand side of our given quadratic equation is a sum or difference of reciprocals. The sum could be something as simple as x + 1/x, or it could be 2x + 3 + 1/(2x + 3) or it could even be (3x + 4)/(4x + 3) – (4x + 3)/(3x + 4). The actual numerators and denominators of the two terms are immaterial as long as they are linear, the two terms are reciprocals of each other, and they are added to each other (there is a “+” sign on the left-hand side between the two terms) or subtracted from each other (there is a “-” sign on the left-hand side between the two terms).

Let us also assume that the right-hand side is a single term (we will consider fractions to be a single term, even though they are actually composed of 2 numbers). If it was already expressed as the sum or difference of two reciprocals, then we would have been done. If it was not expressed in a form similar to the left-hand side (either sum of reciprocals if the left-hand side is a sum, or a difference of reciprocals if the left-hand side is a difference), then we will assume that we have done the necessary arithmetic to convert the right-hand side to a single term. Also let us assume that mixed fractions have been converted to improper fractions if necessary. Thus, our quadratic equation looks as below:

(ax + b)/(cx + d) + (cx + d)/(ax + b) = e or
(ax + b)/(cx + d) – (cx + d)/(ax + b) = e

We already know that if e can be expressed as (f/g + g/f) (or (f/g – g/f) as the case may be), then our task is done. All we need to do at that point is to equate (ax + b)/(cx + d) = f/g and (ax + b)/(cx + d) = g/f (or -g/f as the case may be) to get the two solutions to our quadratic equation. Therefore, we will concentrate our efforts purely on expressing e as (f/g + g/f) or (f/g – g/f).

Note that f/g + g/f is actually equal to (f^2 + g^2)/fg. That gives us the first clue about how to proceed in this case. Let us first consider a case where e is actually a fraction with a numerator and a denominator. Thus e is expressed as h/j. And so, h/j = (f^2 + g^2)/fg.

Why is the above observation significant? Let us take a fraction such as 61/30. At first sight, it looks like it would be very difficult to break it down into the sum of two reciprocals. But consider the fact that h = 61 and j = 30. From the last line of the last paragraph, j is also equal to fg. Therefore, let us find all the pairs of factors for 30. They are:

1 x 30
2 x 15
3 x 10 and
5 x 6

The numerator, h is equal to 61. But it is also equal to (f^2 + g^2). Let us calculate (f^2 + g^2) for each of these pairs of factors above:

1 x 30 : 1 + 900 = 901
2 x 15 : 4 + 225 = 229
3 x 10 : 9 + 100 = 109
5 x 6 : 25 + 36 = 61

We find that our actual numerator, 61, is actually equal to (f^2 + g^2) for one of the pairs of factors of the denominator. Thus 61/30 = (5^2 + 6^2)/(5 x 6). This then means that 61/30 = 5/6 + 6/5. Thus, we have found the pair of reciprocals for this value of e. This then leads to the solution of quadratic equation much quicker than if we had started out by cross-multiplying, collecting the terms and doing the other operations required to get the equation into standard form.

What happens if our right-hand side was actually written as 122/60? We know that it is exactly equal to 61/30, but suppose we work with 122/60 as given. Then what happens? Let us go through the same exercise as above and identify the sum of squares of all the pairs of factors for 60. We get:

1 x 60 : 1 + 3600 = 3601
2 x 30 : 4 + 900 = 904
3 x 20 : 9 + 400 = 409
4 x 15 : 16 + 225 = 241
5 x 12 : 25 + 144 = 169
6 x 10: 36 + 100 = 136

We see that none of the pairs of factors actually gave us a sum of squares equal to 122, our numerator. Here is the problem: suppose the denominator, when the fraction is expressed in lowest terms, is factorable into a pair of factors whose sum of squares is equal to the numerator. Then if both the numerator and denominator are multiplied by a whole number that is not a perfect square, then the denominator will not be factorable into a pair of factors whose sum of squares is equal to the numerator. That is why it is extremely important for the fraction to be in lowest terms when we attempt to do this.

What happens if the right-hand side happened to be -61/30 instead of 61/30? Actually, we find that -61/30 = -6/5 – 5/6. So, we can say that the right-hand side is the sum of two reciprocals, both of which are negative. It is not as if (f^2 + g^2) = -61 (this can never be true except for complex values of f and g), it is just that (-f^2 – g^2) = -61. This is important to remember!

Thus, the steps required to convert a fractional value on the right hand side (a value with a numerator and denominator) to the sum of a pair of reciprocals is as below:

  1. Express the given fraction in lowest terms
  2. If the given fraction is negative, ignore the negative sign for now
  3. Find all pairs of factors, f and g, for the denominator
  4. Check if (f^2 + g^2) is equal to the numerator for any pair of factors
  5. If so, change e into (f/g + g/f) or (-f/g – g/f) (if the given fraction was negative), and solve the quadratic equation

Similarly, note that f/g – g/f is actually equal to (f^2 – g^2)/fg. Taking a case once again where e is actually a fraction with a numerator and a denominator, e can expressed as h/j. And so, h/j = (f^2 – g^2)/fg.

Now, let us take a fraction such as 33/28. At first sight, this also looks like it would be very difficult to break it down into the difference of two reciprocals. But consider the fact that h = 33 and j = 28. From the last line of the last paragraph, j is also equal to fg. Therefore, let us find all the pairs of factors for 28. They are:

1 x 28
2 x 14
4 x 7
7 x 4
14 x 2 and
28 x 1

It is important to write out the last three pairs in this case because difference is not a commutative operation, so we need to check both sides of each pair when doing the subtractions.

The numerator, h is equal to 33. But it is also equal to (f^2 – g^2). Let us calculate (f^2 – g^2) for each of these pairs of factors above:

1 x 28 : 1 – 784 = -783
2 x 14 : 4 – 196 = -192
4 x 7 : 16 – 49 = -33
7 x 4 : 49 – 16 = 33
14 x 2 : 196 – 4 = 192
28 x 1 : 784 – 1 = 783

We find that our actual numerator, 33, is actually equal to (f^2 – g^2) for one of the pairs of factors of the denominator. Thus 33/28 = (7^2 – 4^2)/(7 x 4). This then means that 33/28 = 7/4 – 4/7. Thus, we have found the pair of reciprocals for this value of e. This can then lead to the solution of the quadratic equation much quicker than if we had started out by cross-multiplying, collecting the terms, and doing the other operations required to get the equation into standard form.

The same caution I mentioned earlier about making sure the fraction is in lowest terms applies here also. If the fraction is actually a difference of reciprocals, but its numerator and denominator are multiplied by a common factor that is not a perfect square, then we will face difficulties trying to express it as a difference of reciprocals. So, we have to make sure that the fraction is in lowest terms before we set out on this procedure.

Thus, the steps required to convert a fractional value on the right hand side (a value with a numerator and denominator) to the sum of a pair of reciprocals is as below:

  1. Express the given fraction in lowest terms
  2. Find all pairs of factors, f and g, for the denominator
  3. Check if f^2 – g^2 is equal to the numerator for any pair of factors
  4. If so, change e into f/g – g/f and solve the quadratic equation

Notice that in this case, we don’t make any special provision for negative right-hand sides. For instance if, instead of 33/28, our right-hand side had actually been -33/28, we would have simply found our difference of reciprocals to be 4/7 – 7/4.

For some examples of the applications of these techniques to the solution of actual quadratic equations, please read the full lesson here.

We still have not touched upon the question of what happens if we are not able to split the right-hand side into a pair of reciprocals using the method explained here. Perhaps, the number on the right hand side is a whole number. In that case the denominator is 1, and 1 has only one factor, 1 itself. Expressing 1 as 1 x 1, we can calculate the sum of the squares of the factors as 2. The difference of the squares of the factors is 0.

What this means is that if the right-hand side is 2 and the left-hand side is a sum, the required sum of reciprocals on the right-hand side is 1/1 + 1/1. Similarly, if the right-hand side is zero and the left-hand side is a difference, the required difference of reciprocals on the right-hand side is 1/1 – 1/1.

Unfortunately, for any other whole number, things are not as easy as this. And the same is true for many fractions whose denominators don’t have factors whose squares add up to the required numerator (or can give the required numerator by subtracting, as the case may be). We will leave these problems to deal with in the next lesson. In the meantime, good luck, and happy computing!

– The Vedic Maths Forum India


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Gaurav Tekriwal is the founder President of the Vedic Maths Forum India. Through television programs, workshops, DVDs, and Books he has taken the Vedic Maths System to over 4 million students in India, South Africa, United States, Australia, UAE, Ghana, and Colombia.

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