## Wednesday, February 10, 2010

### Vedic Mathematics Lesson 39: Quadratic Equations 3

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.

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

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.
`RightHand    Reciprocal1    Reciprocal2Side-----------------------------------3        2.6180         0.38204        3.7321         0.26795        4.7913         0.21876        5.8284         0.17167        6.8541         0.14598        7.8730         0.12709        8.8875         0.112510        9.8990         0.101011       10.9083         0.091712       11.9161         0.083913       12.9226         0.077414       13.9282         0.071815       14.9330         0.067016       15.9373         0.062717       16.9410         0.059018       17.9443         0.055719       18.9472         0.052820       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.

Let us solve a few examples that fall under the purview of type 1 so that we understand how to apply the method quickly and correctly. First, let us solve the equation below:

x + 1 + 1/(x + 1) = 5

The left hand side is a sum of reciprocals, and the right hand side is a whole number. So, the equation does fall under the purview of case 1. Moreover, fortunately the right hand side is not -1, 0 or 1, so the roots of the equation are not complex. Unfortunately, the right hand side is not -2 or 2, so the reciprocals are not 1/1 and 1/1.

To solve this equation, we can either use the exact reciprocals we have derived in the table above, or we can choose to use e - 1/e and 1/e as the two reciprocals. Using the exact reciprocals from the table above, we set x + 1 to be equal to 4.7913 for one root of 3.7913. We then set x + 1 to be equal to 0.2187 for the other root of -0.7913. If we had used e - 1/e and 1/e as the two reciprocals, we would have calculated the roots as 3.8 and -0.8. As you can see we would not have been very far off the real solutions!

Next, let us tackle:

(4x + 7)/(2x + 3) + (2x + 3)/(4x + 7) = -50

In this case, after verifying that the equation is of type 1, and that the value of e is well outside the -2 to 2 range, we can decide to express -50 as reciprocals using the approximation of e - 1/e and 1/e. Notice that the value of e is negative, so we get the reciprocals as -49.98 and -0.02. Setting (4x + 7)/(2x + 3) to equal -49.98 gives us a root of approximately -1.5 and setting (4x + 7)/(2x + 3) = -0.02 gives us the other root of approximately -1.75. The actual roots, if you care to work them out after expanding the given equation out to the standard form and applying the quadratic formula are -1.5096 and -1.7475. As you can see, given that this technique is all about quick, mental calculations, we did not do too badly on an equation that would take longer to get into the standard form than we spent solving it to 2 decimal places correctly!

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

• 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.
`RightHand    Reciprocal1    Reciprocal2Side-----------------------------------1        1.6180          0.61802        2.4142          0.41423        3.3028          0.30284        4.2361          0.23615        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.

Let us solve a few examples that fall under the purview of case 2 so that we understand how to apply the method quickly and correctly. First, let us solve the equation below:

x - 1/x = 3

The left hand side is a difference of reciprocals, and the right hand side is a whole number. So, the equation does fall under the purview of case 2. Moreover, unfortunately the right hand side is not 0, so the reciprocals are not 1/1 and 1/1.

We can use the table above to express the equation above as:

x - 1/x = 3.3028 - 0.3028

This immediately gives us the solutions x = 3.3028 and x = -0.3028.

Now, let us tackle:

(3x + 5)/5x - 5x/(3x + 5) = -10

We verify that this indeed an equation of type 2. Instead of calculating exact values, we will use the approximations of e + 1/e and 1/e as the reciprocals to use in this case. Since the right hand side is negative, we can rewrite the equation above to:

(3x + 5)/5x - 5x/(3x + 5) = 0.1 - 10.1

Setting (3x + 5)/5x = 0.1 gets us the solution x = -2. Setting (3x + 5)/5x = -10.1 gives us the other solution, x = -0.1 The exact solutions turn out to be -1.9961 and -0.0935 (rounded off to 4 digits after the decimal point). Once again, not too bad given that we are able to solve the equation in much less time than it takes simply to get it into the standard form in preparation for applying the quadratic formula!

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!

For instance, consider the equation:

x + 1/x = 22/3

Since 22/3 is over 5, we know that the approximate reciprocals work about as well as the true reciprocals. So, we can simply express 22/3 as 22/3 - 3/22 + 3/22, which then gives us 475/66 + 3/22. The roots of our equation are thus 475/66 and 3/22. Expressing them in decimal form, we get x = 7.1970 and 0.1364. The exact solutions to the equation turn out to be x = 7.1943 and 0.1390. Pretty close, if I do say so myself!

Moreover, the right hand side need not even be in the form of a fraction. We can use our approximations to find pairs of reciprocals even when the right hand side is in the form of decimal number. Consider the equation:

(x + 1)/(x + 2) + (x + 2)/(x + 1) = -7.256

Since the absolute value 0f -7.256 is over 5, we can use our approximation without being too far off. So, we express -7.256 as -7.256 + 1/7.256 - 1/7.256. Setting the value of (x + 1)/(x + 2) to -7.256 + 1/7.256, we still are faced with some daunting squaring and dividing, but these are nowhere near as complicated as finding multiple square roots. We finally get the root as x = -1.8768. Setting (x + 1)/(x + 2) to -1/7.256, we get the other root as x = -1.1211. The true roots of the equation turn out to be x = -1.8768 and -1.1232. Once again, pretty close!

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!

For instance, consider the equation:

x - 1/x = 33/4

Since 33/4 is over 5, we know that the approximate reciprocals work about as well as the true reciprocals. So, we can simply express 33/4 as 33/4 + 4/33 - 4/33, which then gives us 1105/132 - 4/33. The roots of our equation are thus 1105/132 and -4/33. Expressing them in decimal form, we get x = 8.3712 and -0.1212 The exact solutions to the equation turn out to be x = 8.3695 and -0.1195. Pretty close, if I do say so myself!

Just as in the case of type 3 equations, we can deal with non-fractional right hand sides in type 4 equations also (right hand sides that are expressed in the form of decimals). Take the following equation, for instance:

(x + 1)/(x + 2) - (x + 2)/(x + 1) = 6.283

Since the absolute value of 6.283 is over 5, we will use approximate reciprocals rather than bothering with several squares and square roots. We can express 6.283 as 6.283 + 1/6.283 - 1/6.283. We then set (x + 1)/(x + 2) equal to 6.283 + 1/6.283 to get one root, x = -2.1838. We then set (x + 1)/(x + 2) equal to 1/6.283 to get the other root, x = -1.1373. The exact roots of the equation turn out to be x = -2.1839 and x = 1.1344.

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!

#### 1 comment:

Anonymous said...

only first quadratic equation 1 rocks
but other post seem to have a headache ,worst than traditional method .

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