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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 saw how it is possible to express some fractions as sums or differences of reciprocals. This is very handy in solving some quadratic equations that are expressed as sums or differences of reciprocal linear expressions involving the unknown on one side, and sums or differences (respectively) of constants on the other side of the equal to sign. In particular, the technique involved factorizing the denominator of the constant term on the right hand side of the equation, and then finding the set of factors whose sum or difference of squares equalled the numerator.

However, we also saw that it may not always be possible to express the constant term as a sum or difference of reciprocals. As we mentioned in the previous lesson, this is true for most whole numbers as well as numerous fractions. This lesson will go into some details of how to deal with such right hand sides.

For the purposes of this lesson, we will assume that the left hand side of our quadratic equation is similar to the left hand side we have been dealing with in the past couple of lessons: it is either a sum or reciprocals or a difference of reciprocals. It can be as simple as x + 1/x or as complicated as (2x + 5)/(3x + 2) – (3x + 2)/(2x + 5).

Now let us look at the right hand side. Let us assume that the right hand side was not capable of being expressed as the sum or difference of reciprocals as explained in the previous lesson. What do we do in that case?

In this lesson, we will handle this as four separate types of cases. In the first type, the left hand side is a sum of reciprocals and the right hand side is a whole number. In the second type, the left hand side is a difference of reciprocals and the right hand side is a whole number. In the third type, the left hand side is once again a sum of reciprocals, but the right hand side is a fraction. Similarly, in the fourth type, the left hand side is a difference of reciprocals, and the right hand side is a fraction. Note that in all these types, in most cases, the roots of the given quadratic equations will be at best irrational, and at worst complex numbers.

Type 1: Left hand side is a sum of reciprocals. Right hand side is a whole number.

The equation we have can be expressed in general terms as below:

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

In this type, e is a whole number that can be expressed as e/1 in its lowest terms. A few observations can be made immediately:

• No number between -2 and +2 (not inclusive) can be expressed as the sum of reciprocals. Therefore, if the number e is either -1, 0 or 1, then the equation has no real roots. It has only complex roots, which we will consider to be outside the scope of this lesson
• -2 and +2 can be expressed as sums of reciprocals by simple examination. -2 = -1/1 + 1/-1. Similarly, 2 = 1/1 + 1/1. These are the only two values of the right hand side that give us rational roots for the equation
• Any solution we find for a positive e will hold for the same negative e with the signs on the reciprocals changed from positive to negative (as we saw in the previous lesson)

Given the third observation above, we will concentrate on positive values of e, knowing that the same solution is applicable to a negative value of e with a change in sign.

If e is to be expressed as the sum of reciprocals, f/g + g/f, we see that f^2 + g^2 = e and fg = 1. Solving these two equations, and substituting the solutions back in f/g + g/f, we can see that the following representation of e will allow us to solve our equation:

e = sqrt(e + sqrt(e^2 – 4))/sqrt(e – sqrt(e^2 -4)) + sqrt(e – sqrt(e^2 -4))/sqrt(e + sqrt(e^2 – 4))

I have used sqrt() to denote square roots rather than drawing in square root symbols! Notice that if you add up the two fractions above, you will get 2*e in the numerator and 2 in the denominator, giving us e as the final result. We can now equate each of the terms on the left hand side to either of the numbers above to get our final solutions.

Thus, for instance, if the right hand side were 7, it would be expressed as sqrt(7 + sqrt(45))/sqrt(7 – sqrt(45)) + sqrt(7 – sqrt(45))/sqrt(7 + sqrt(45)). If you actually work out the square roots in the expression above, you would get the expression below (I have rounded off the actual results, which are irrational numbers with infinitely many digits after the decimal point, to 4 digits after the decimal point):

7 = 6.8541 + 0.1459

Obviously, it easy to verify that the sum of the two is indeed 7. But I encourage you to verify that the two numbers are indeed reciprocals of each other also!

As you can see, though, the final solution looks quite daunting. If the left hand side is simple (for instance, it is just x + 1/x), we may be better off expanding out the terms and applying the quadratic formula (which by the way will give us the exact same solutions as we have derive using this methodology). However, if the left hand side is more complicated, it may still be easier to apply this method to solve it rather than expanding out the left hand side, cross-multiplying and finally getting the equation into standard form.

To aid in this, I have provided a table below that lists the actual values of the two reciprocals (rounded off to 4 digits after the decimal place) that add up to e for different values of e from 3 through 20. If you have a good memory, you may be able to keep at least the approximate values of these reciprocals in mind, and when you encounter an equation that fits the mold, you may be able to solve it mentally without having to use the quadratic formula.

```Right
Hand    Reciprocal1    Reciprocal2
Side
-----------------------------------
3        2.6180         0.3820
4        3.7321         0.2679
5        4.7913         0.2187
6        5.8284         0.1716
7        6.8541         0.1459
8        7.8730         0.1270
9        8.8875         0.1125
10        9.8990         0.1010
11       10.9083         0.0917
12       11.9161         0.0839
13       12.9226         0.0774
14       13.9282         0.0718
15       14.9330         0.0670
16       15.9373         0.0627
17       16.9410         0.0590
18       17.9443         0.0557
19       18.9472         0.0528
20       19.9499         0.0501```

As you can see, the two reciprocals approach e – 1/e and 1/e for even modestly large values of e. In fact, when e becomes 100, the two reciprocals, rounded off to 5 digits after the decimal place, are in fact, e – 1/e and 1/e! Thus you can use those values in your calculations in a pinch!! So, we may not be as badly off as we feared even if we don’t have good memories.

In the interest of keeping this lesson short, I have not provided examples of how to use the methods in this lesson. But, if you are interested, please read the full post on my blog here. You may be surprised at how much labor-savings this solution method can yield!

Type 2: Left hand side is a difference of reciprocals. Right hand side is a whole number.

Let us start with a couple of simple observations, once again.

• All real numbers can be expressed as a difference of reciprocals without the use of complex numbers
• In particular, 0 can be expressed as 1/1 – 1/1, so if the right hand side is zero, we are already done

Once again, as with type 1 equations, type 2 equations can be expressed as:

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

If e is to be expressed as the difference of reciprocals, f/g – g/f, we see that f^2 – g^2 = e and fg = 1. Solving these two equations, and substituting the solutions back in f/g – g/f, we can see that the following representation of e will allow us to solve our equation:

e = sqrt(sqrt(e^2 + 4) + e)/sqrt(sqrt(e^2 + 4) – e) – sqrt(sqrt(e^2 + 4) – e)/sqrt(sqrt(e^2 + 4) + e)

Once again, I have used sqrt() to denote square roots rather than drawing in square root symbols! Notice that if you add up the two fractions above, you will get 2*e in the numerator and 2 in the denominator, giving us e as the final result.

Thus, for instance, if the right hand side were 7, it would be expressed as sqrt(sqrt(53) + 7)/sqrt(sqrt(53) – 7) – sqrt(sqrt(53) – 7)/sqrt(sqrt(53) + 7) . If you actually work out the square roots in the expression above, you would get the expression below (I have rounded off the actual results, which are irrational numbers with infinitely many digits after the decimal point, to 4 digits after the decimal point):

7 = 7.1401 – 0.1401

Obviously, it easy to verify that the difference of the two is indeed 7. But I encourage you to verify that the two numbers are indeed reciprocals of each other also!

I have provided the values of these reciprocals for the first few whole numbers in the table below (rounded off to 4 digits after the decimal place). Note that the smaller reciprocal is subtracted from the larger one for positive values of the right hand side, and the larger reciprocal is subtracted from the smaller one for negative values of the right hand side. The table below, will therefore, only list positive values of the right hand side.

```Right
Hand    Reciprocal1    Reciprocal2
Side
-----------------------------------
1        1.6180          0.6180
2        2.4142          0.4142
3        3.3028          0.3028
4        4.2361          0.2361
5        5.1926          0.1926```

You will notice that the values of the reciprocals start to approach e + 1/e and 1/e. Thus, we can use those as approximations rather than calculating the exact values of the reciprocals for larger values of the right hand side.

In the interest of keeping this lesson short, I have not provided examples of how to use the methods in this lesson. But, if you are interested, please read the full post on my blog here. You may be surprised at how much labor-savings this solution method can yield!

Type 3: Left hand side is a sum of reciprocals. Right hand side is a fraction.

The equation we have can be expressed in general terms as below:

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

If e/f is between -2 and +2 (not inclusive), then the solution to the equation is complex and outside the scope of this lesson. We will therefore assume that e/f is either less than -2 or greater than +2 (e/f can not be exactly -2 or +2 because we assume that e/f is a fraction expressed in its lowest terms and is not a whole number). Once again, we will concentrate on positive values, knowing that we can easily derive the solutions for negative values by changing the signs on the reciprocals.

By expressing e/f as (g^2 + h^2)/gh, we can solve for the reciprocals to be used. We find that the reciprocals are as below:

e/f = sqrt(e + sqrt(e^2 – 4*f^2))/sqrt(e – sqrt(e^2 – 4*f^2)) + sqrt(e – sqrt(e^2 – 4*f^2))/sqrt(e + sqrt(e^2 – 4*f^2))

This is obviously quite complicated. For instance, let us take a case where the right hand side is 7/2. In this case, e is 7 and f is 2. Substituting them in the solution above, we find that:

7/2 = sqrt(7 + sqrt(49 – 4*2^2))/sqrt(7 – sqrt(49 – 4*2^2)) + sqrt(7 – sqrt(49 – 4*2^2))/sqrt(7 + sqrt(49 – 4*2^2)) which becomes
7/2 = 3.1861 + 0.3139

Obviously, 3.1861 + 0.3139 = 3.5, which is the same as 7/2. I encourage you to verify that that 3.1861 and 0.3139 are also reciprocals of each other.

Unfortunately, as you can see, this method yields us the result we need, but the computational cost could very well be too high. Fortunately, we can still use the approximation we derived, for finding the reciprocals in type 1, to find the reciprocals in type 3 equations also. In particular, if we find that the absolute value of e/f is 5 or above, we can use e/f – f/e and f/e as the two approximate reciprocals. This will enable us to bypass the complicated procedures involved in deriving the true reciprocals, so we can actually solve such equations mentally rather than requiring reams of paper!

In the interest of keeping this lesson short, I have not provided examples of how to use the methods in this lesson. But, if you are interested, please read the full post on my blog here. You may be surprised at how much labor-savings this solution method can yield!

Type 4: Left hand side is a difference of reciprocals. Right hand side is a fraction.

The equation we have can be expressed in general terms as below:

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

By expressing e/f as (g^2 – h^2)/gh, we can solve for the reciprocals to be used. We find that the reciprocals are as below:

e/f = sqrt(sqrt(e^2 + 4*f^2) + e)/sqrt(sqrt(e^2 + 4*f^2) – e) – sqrt(sqrt(e^2 + 4*f^2) – e)/sqrt(sqrt(e^2 + 4*f^2) + e)

This is obviously quite complicated. For instance, let us take a case where the right hand side is 11/3. In this case, e is 11 and f is 3. Substituting them in the solution above, we find that:

11/3 = sqrt(sqrt(11^2 + 4*3^2) + 11)/sqrt(sqrt(11^2 + 4*3^2) – 11) – sqrt(sqrt(11^2 + 4*3^2) – 11)/sqrt(sqrt(11^2 + 4*3^2) + 11) which becomes
11/3 = 3.9217 – 0.2550

Obviously, 3.9217 – 0.2550 = 3.6667, which is the same as 11/3. I encourage you to verify that that 3.9217 and 0.2550 are also reciprocals of each other!

Unfortunately, as you can see, this method yields us the result we need, but the computational cost could very well be too high. Fortunately, we can still use the approximation we derived, for finding the reciprocals in type 2, to find the reciprocals in type 4 equations also. In particular, if we find that the absolute value of e/f is 5 or above, we can use e/f + f/e and f/e as the two approximate reciprocals. This will enable us to bypass the complicated procedures involved in deriving the true reciprocals, so we can actually solve such equations mentally rather than requiring reams of paper!

In the interest of keeping this lesson short, I have not provided examples of how to use the methods in this lesson. But, if you are interested, please read the full post on my blog here. You may be surprised at how much labor-savings this solution method can yield!

This lesson has already become very long, and I don’t want to make it any longer with a lengthy conclusion. So, for now, good luck, and happy computing!

– The Vedic Maths Forum India

#### 2 COMMENTS

1. HI Vedic Maths!!
Pl let me know how to find 'SQUARE ROOT'
of numbers, the Vedic Way!!

Thanks ,
T.Sasisekaran
[a Senior Citizen, pushing -74years]
15th Feb '10…06:05am

2. Thank you for your request. I will certainly be covering that topic in a future lesson. I am not sure when, but I will certainly keep it in mind, and not skip it. In the meantime, take a look at the Babylonian method for calculating square roots in this wikipedia article. It is very simple and can be done mentally with just a little bit of practice.
The Babylonian method is a very useful method not just to calculate square roots, but also many other solutions. The simple iterative scheme used in the method can be modified and can then be used to solve polynomial equations, solve simultaneous equations, calculate arbitrary roots (not just square roots), etc.