 Namaste. Myself, Dr. Basaraj Emberadhar, Assistant Professor, Department of Communities and Sciences, Valchain Institute of Technology, Zolapur. In this video, I explain the numerical solution of algebraic and transcendental equations by Regula-Falci-Methan. Learning outcomes. At the end of this session, the student will be able to find the roots of algebraic and transcendental equations by Regula-Falci-Methan. Pause the video and write the formula of Regula-Falci-Methan. I hope all of you have written the answer. Answer. Suppose f of x is equal to 0 and f of a and f of b are opposite signs, then the equation f of x is equal to 0 will have at least one real root between the interval a, b and is given by x is equal to a into f of b minus b into f of a, 4 divided by f of b minus f of a. This is known as Regula-Falci formula. Come to an example. Use the Regula-Falci-Methan to find the real root of the equation x cube minus 2x minus 50 is equal to 0 correct to 3 decimal of places. Let f of x is equal to x cube minus 2x minus 50 to find the interval a, put x is equal to 0, 1, 2, 3 and so on, where the sign of f of x changes either from positive to negative or from negative to positive 2, then the corresponding value of x is taken as an even interval. For the same f of 0 is equal to, put x is equal to 0 in the given equation that is minus 50 which is negative and f of 1 is equal to minus 51 which is also negative, f of 2 which is equal to minus 46 which is also negative and f of 3 is equal to minus 29 which is also negative and f of 4 is equal to 6 which is a positive that is greater than 0. It is observed that the value of f of x at x is equal to 4 is being 6 is in error 2, 0 as compared to f of 3 is equal to minus 29 and we expect the root in the neighbor of 4. Now, for that we have to substitute the between the value of 3 and 4 here that is f of 3.9 near to 4 that is 3.9 which is equal to 1.5119 which is greater than 0 and f of 3.8 is equal to minus 2.728 which is the negative. Since f of 3.8 is equal to minus 2.728 which is less than 0 and f of 3.9 is equal to 1.5119 which is greater than 0 such that f of 3.8 into f of 3.9 is less than 0. Therefore, the root lies in the interval 3.8 and the 3.9. Then by the regular falsification law is the x is equal to a into f of b minus b into f of a whole divided by f of b minus f of a call it is equation 1. Come to an first iteration here the a is equal to 3.8 and b is equal to 3.9 and the f of a is equal to minus 2.728 and f of b is equal to 1.5119 substituting these values in equation 1 and simplifying we get that is x 1 indicates that is the approximation that is x 1 is equal to 3.8643. Now substitute the value of x is equal to x 1 is equal to 3.8643 in f of x that gives the f of 3.8643 is equal to minus 0.0237. Since f of 3.8643 is equal to minus 0.237 which is less than 0 and f of 3.9 is equal to 1.5119 which is greater than 0 such that f of 3.8643 into f of 3.9 is less than 0. Therefore, the root lies in the interval 3.8643 comma 3.9. Here a is equal to 3.8643 and b is equal to 3.9 and the f of a is equal to minus 0.0237 and f of b is equal to 1.5119 substitute these values in equation 1 and simplifying we get x 2 is equal to 3.8648. Now comparing the x 1 and x 2 are same up to 3 decimal of places. Hence the approximate root of the given equation is x is equal to 3.8648 correct 2 3 decimal of places. Come to another example use the regula falsi method to find the root of the equation x log x to the base 10 is equal to 1.2 correct 2 4 decimal of places. Solution let f of x is equal to x log x to the base 10 minus 1.2. First we have to find the interval now by putting x is equal to 1 2 3 and so on with the sign of f of x changes either from positive to negative or from negative to positive then the corresponding values of x are to be taken as your interval that is f of 1 is equal to minus 1.2 which is less than 0 that is negative f of 2 is equal to minus 0.5979 which is also less than 0 and f of 3 is equal to 0.231 which is greater than 0. Since f of 2 is less than 0 and f of 3 is greater than 0 such that f of 2 into f of 3 is less than 0. Therefore the root lies in the interval 23. It is observed that the value of f of x at x is equal to 3 is being the 0.231 is nearer to 0 as compared to f of 2 is equal to minus 0.5919 and we expect the root near the neighbor of 3. Now to find the closure interval here we have to substitute the between the value of 2 and 3 that is f of 2.1 f of 2.2 f of 2.3 and so on where the sign of f of x changes either from negative to positive or from positive to negative the same procedure that is the f of 2.7 is equal to minus 0.035 which is less than 0 and f of 2.8 is equal to 0.052 which is greater than 0. Since the f of 2.7 is equal to minus 0.035 which is less than 0 and f of 2.8 is equal to 0.052 is greater than 0 such that f of 2.7 into f of 2.8 is less than 0. Therefore the root lies in the interval 2.7 comma 2.8. The regular falsely formulae is x is equal to a into f of b minus b into f of a 4 divided by f of b minus f of a call it is equation 1. First iteration here a is equal to 2.7 and b is equal to 2.8 and f of a is equal to minus 0.035 f of b is equal to 0.052. Substituting these values in equation 1 and simplifying we get x1 is equal to 2.7402 up to the value of x is equal to x1 is equal to 2.7402 in f of x that is f of 2.7402 is equal to minus 0.0038. Since f of 2.7402 is equal to minus 0.0038 which is less than 0 and f of 2.8 is equal to 0.052 which is greater than 0 such that f of 2.7402 into f of 2.8 is less than 0. Therefore, the root lies in the interval 2.7402 comma 2.8. Second iteration here a is equal to 2.7402 and b is equal to 2.8 f of a is equal to minus 0.00038 and f of b is equal to 0.052. Substitute these values in equation 1 and simplifying we get x2 is equal to 2.7406. Substitute the value of x is equal to x2 is equal to 2.7406 in f of x that is f of 2.7406 is equal to minus 0.00004. Since f of 2.7406 is equal to minus 0.00004 is less than 0 and f of 2.8 is equal to 0.052 is greater than 0 such that f of 2.7406 into f of 2.8 is less than 0. Therefore, the root lies in the interval 2.7406 comma 2.8. Third iteration here a is equal to 2.7406 and b is equal to 2.8 f of a is equal to minus 0.00004 and f of b is equal to 0.052. Substitute these values in equation 1 and simplify we get x3 is equal to 2.7406. Comparing the x2 and x3 we observe that they are same up to 4 decimal of places. Hence, the approximate root of the given equation is x is equal to 2.7406 correct to 4 decimal of places. References. Thank you.