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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! For those who read the first part of my story, I have posted the second part 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 past several lessons, we have dealt with the solution of various types of linear equations. We started out by deriving various formulae to solve equations that fall into specific types using the Paravartya Yojayet sutra. We then proceeded to applications of the powerful Sunyam Samyasamuccaye sutra. We then covered various kinds of mergers, and finally we covered applications of the Anurupye Sunyam Anyat sutra to the solution of simultaneous linear equations.

In this lesson, we will deal with quadratic equations, which are polynomial equations of the second degree. What this means is that when the equation is expanded out with no fractional terms, there exists at least one term in which the unknown quantity is raised to the second power.

The general form of a quadratic equation is ax^2 + bx + c. The quadratic formula can be used to solve equations in the standard form. The quadratic formula is usually written as below:

quadratic formula
In this lesson, however, we will not deal with quadratic equations that are written in the standard form. It may take quite a bit of effort in cross-multiplication, collection of like terms, etc., to get these equations into the standard form. The solution of these equations using the quadratic formula, therefore, is quite cumbersome, not to mention, error-prone.

Rather than go through all that labor, we will identify these special types of equations and apply certain simple procedures to solve them quickly and easily.

The first special type of quadratic equations we will consider are like the one below:

x + 1/x = 10/3

In the traditional method, we would go through the process of getting the equation into the standard form using the steps below:

x + 1/x = 10/3 becomes
(x^2 + 1)/x = 10/3 becomes
3x^2 + 3 = 10x becomes
3x^2 – 10x + 3 = 0

We would then solve it using the quadratic formula by setting a = 3, b = -10 and c = 3. This would lead to the solutions x = 3 and x = 1/3.

However, we don’t have to go through all that trouble to solve this kind of equation. All we have to do is observe that 10/3 = 3 + 1/3. Thus, we would immediately have figured out that we can rewrite the given equation as below:

x + 1/x = 3 + 1/3

The symmetry of the equation above immediately reveals the answers, x = 3 and x = 1/3.

The same method can be used to solve many different problems such as the examples below:

x + 1/x = 26/5 => x + 1/x = 5 + 1/5 => x = 5, 1/5
x + 1/x = 50/7 => x + 1/x = 7 + 1/7 => x = 7, 1/7
x + 1/x = -17/4 => x + 1/x = -4 – 1/4 => x = -4, -1/4

We are not restricted to x + 1/x on the left-hand side either. Consider the equation:

(2x + 3) + 1/(2x + 3) = 50/7

Since the right-hand side of the equation can be expanded to 7 + 1/7, by the symmetry of the equation, we can equate 2x + 3 to either 7 or 1/7 (or equivalently, by equating (2x + 3) and 1/(2x + 3) to 7). We then get the solutions x = 2 and x = -10/7 to the given equation.

Similarly, consider the equation:

5x/(2x + 3) + (2x + 3)/5x = 26/5

The right-hand side of the equation can be expanded to 5 + 1/5. Thus, we can use the symmetry of the resulting equation to derive the following linear equations:

5x/(2x + 3) = 5
5x/(2x + 3) = 1/5 or alternatively, (2x + 3)/5x = 5

These two equations can then be solved to give us x = -3 or x = 23/3.

Now, consider the equation:

(x + 3)/(3x + 5) + (3x + 5)/(x + 3) = 17/4

The right-hand side of the equation can be expanded to 4 + 1/4. This then lets us solve the equation by deriving the linear equations below:

(x + 3)/(3x + 5) = 4
(x + 3)/(3x + 5) = 1/4 or alternatively, (3x + 5)/(x + 3) = 4

These equations can then be solved to give us x = -7 and x = -17/11.

Now, consider an equation of the type below:

x – 1/x = 3/2

We can rewrite the equation as below:

x – 1/x = 2 – 1/2

We may be tempted to conclude from the symmetry of the equation on both sides of the equal-to sign that x = 2 or x = 1/2. That would be wrong. In equations such as the above where the terms are connected by “-” signs instead of “+” signs, the solutions are x = 2 and x = -1/2. Only with x = -1/2 is it possible to get -1/x = 2, and therefore x – 1/x = 2 – 1/2. This is important to remember.

We will illustrate this with a few examples as below:

x – 1/x = 8/3 becomes
x – 1/x = 3 – 1/3, which then leads to x = 3 and x = -1/3 as the solutions.

x – 1/x = 63/8 becomes
x – 1/x = 8 – 1/8, which then leads to x = 8 and x = -1/8 as the solutions.

x – 1/x = -24/5 becomes
x – 1/x = -5 + 1/5 which then leads to x = -5 and x = 1/5 as the solutions.

The technique is equally applicable to cases where the left-hand side consists of other terms than x and 1/x. The following examples illustrate a few examples of these cases:

(3x + 2) – 1/(3x + 2) = 63/8 => (3x + 2) – 1/(3x + 2) = 8 – 1/8 =>
3x + 2 = 8, 3x + 2 = -1/8 => x = 2, x = -17/24
2x/(5x + 1) – (5x + 1)/2x = -15/4 => 2x/(5x + 1) – (5x + 1)/2x = 1/4 – 4 =>
2x/(5x + 1) = 1/4, 2x/(5x + 1) = -4 => x = 1/3, -2/11
(4x + 3)/(3x + 4) – (3x + 4)/(4x + 3) = 24/5 => (4x + 3)/(3x + 4) – (3x + 4)/(4x + 3) = 5 – 1/5 =>
(4x + 3)/(3x + 4) = 5, (4x + 3)/(3x + 4) = -1/5 => x = -17/11, -19/23

Sometimes, the equation may have undergone some transformations that hide its true nature. And sometimes, the right-hand side looks as if it is not really the sum or difference of reciprocals. But closer observation leads to the solution by unmasking the true nature of the equation. For some examples of these kinds of problems, please read the full lesson here.

Assuming that the left-hand side is a pair of reciprocals, connected by either “+” or “-“, how do we verify whether the right-hand side can be expressed as a pair of reciprocals with the same sign between them? We will explore this question in greater detail in the next lesson.

In the meantime, I hope you have found this lesson useful and interesting. I also hope you will apply the techniques explained in this lesson on real problems so that you become familiar not only with the technique itself (which is actually quite trivial), but also with the fractions that result either from the addition of numbers with their reciprocals, or the differences between numbers and their reciprocals. That will enable one to apply this technique where appropriate, on sight and mentally, to solve the types of quadratic equations we have dealt with in this lesson. 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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