 Namaste. Myself, Dr. Basaraj Ambiradhar, Assistant Professor, Department of Humanities and Sciences, Valchain Institute of Technology, Solapur. In this video, I explain the numerical solution of algebraic and transcendental equation by Regalof Valchi method. Learning Outcomes At the end of this session, the student will be able to find the roots of algebraic and transcendental equation by Regalof Valchi method. In the field of the science and engineering, the solutions of the form f of x is equal to 0 occurs in many applications. If f of x is the polynomial of degree 2 like x square plus 3 x plus 3 is equal to 0 or the degree 3 or degree 4 exact solutions are available. But, if f of x is a transcendental function like, means the equation which contains the logarithms or the trigonometric or exponential and algebraic, then such equations are called a transcendental equation. For example, 3 x plus log x is equal to 7, the e to the power minus x is equal to sin x and etcetera. The solution is not exact, but do not have formulae to get the solution. In such a situation, we adopt the numerical approximate method to solve the such algebraic and the transcendental equations. If f of x is continuous in the interval a, b and f of a and f of b have opposites, then the equation f of x is equal to 0 has at least one root between x is equal to a and x is equal to b. Now, find the interval where the root of the equation x cube minus 2 x minus 5 is equal to 0 is located solution. Let f of x is equal to x cube minus 2 x minus 5. Now, to find the interval, we have to put x is equal to 0, 1, 2, 3 and so on. Where the sin of f of x changes either from positive to negative or from negative to positive, then the corresponding value of x is considered or to be taken as an even interval. For that f of 0 is equal to substituting that is equal to minus 5 which is less than 0 and f of 1 is equal to minus 6 which is less than 0 and f of 2 is equal to minus 1 which is less than 0 that is negative a and f of 3 is equal to 16 which is greater than 0. Since, the f of 2 is equal to minus 1 which is less than 0 and f of 3 is equal to 16 which is greater than 0, such that f of 2 into f of 3 is less than 0. Therefore, the root lies in the interval 2, 3. Pause the video and find the interval where the root of the equation x cube minus 9 x plus 1 is equal to 0 is located. I hope all of you return the answer. Solution, let f of x is equal to x cube minus 9 x plus 1. To find the interval, we have to substitute x is equal to 0, 1, 2, 3 and so on, where the f of x sign of the f of x changes either from positive to negative or from negative to positive. Then the corresponding value of x are to be considered as an even interval. That is f of 0 is equal to 1 which is greater than 0. That is positive and f of 1 is equal to minus 7 which is less than 0. Since f of 0 is equal to 1 greater than 0 and f of 1 is equal to minus 7 which is less than 0, such that f of 0 into f of 1 is less than 0. Therefore, the root lies in the interval 0, 1. Now, there are so many methods to find the roots of such algebraic and transcendental equation. Now, today I explain one of the method that is the regular falsely method or it is also called method of falsely position. Consider the equation f of x is equal to 0 and let f of a and f of b be opposite signs and a is less than b. Now, the curve y is equal to f of x in the interval a, b crosses the x axis only at the point p as shown in the figure. That is it crosses the point at the point p which has shown in the figure here. This curve or crosses the point crosses the x axis at a point p here as shown in the figure that a b a of a comma f of a and b of b comma f of b be any two point on the curve y is equal to f of x. We know that the equation of the line joining the two points at a of x 1 y 1 comma b of x 2 y 2 is y minus y 1 upon x minus x 1 is equal to y 2 minus y 1 upon x 2 minus x 1. Therefore, the equation of the curve a b as shown in the figure is y minus f of a upon x minus a is equal to f of b minus f of a whole divided by b minus a. The curve a b crosses the x axis we must have y is equal to 0 at p dash then 0 minus f of a upon x minus a is equal to f of b minus f of a divided by b minus a. On the cross multiplication here x minus a into f of b minus f of a is equal to minus b minus a into f of a. Now, multiply x and a by f of b minus f of a separately and taking on the except the terms which contains x on the left side you have to keep and the terms which not contain x are the other side here that is x into f of b minus f of a which is equal to a into f of b minus a into f of a minus b into f of a plus a into f of a that is plus a into f of a and minus a into f of a get cancelled that is the x is equal to that is r x 1 is equal to x 1 indicates that is the first it is in here that is a into f of b minus b into f of a whole divided by f of b minus f of a. This value of x 1 gives the approximate value of the root f of x is equal to 0 that is a is the less than x less than b it means the value of x 1 is lies between the a and b here and it is called the first iteration or approximation substitute x is equal to x 1 in f of x if f of x 1 is equal to 0 then x 1 is the root of the equation. If f of x 1 is not equal to 0 then check it sign now f of x 1 and f of a are of opposite sign or f of x 1 and f of b are of opposite sign if f of x 1 into f of a is less than 0 or f of x 1 into f of b is less than 0 then x 2 is lies between x 1 and a or x 1 and b respectively we get x 2 is equal to a into f of x 1 minus x 1 into f of a whole divided by f of x 1 minus f of a or x 2 is equal to b into f of x 1 minus x 1 into f of b whole divided by f of x 1 minus f of b it is called the second iteration or approximation continue the process of iteration till we get the desired order of accuracy this process of finding the roots is called the regular falsi method or method of falsi position references thank you