 Namaste. Myself, Dr. Basaraj Emberadhar, Assistant Professor, Department of Humanities and Sciences, Valchain Institute of Technology, Solapur. In this video, I explain the numerical solution of algebraic and transcendental equations by Raghlok-Valchi method, Learning Outcomes. At the end of this session, the student will be able to find the roots of algebraic and transcendental equations by Raghlok-Valchi method. Pause the video and answer the question. Find the interval where the root of the equation 2x minus log x to the base 10 minus 7 is equal to 0 is located. I hope all of you return the answer. Solution. Let f of x is equal to 2x minus log x to the base 10 minus 7. Now, to find the interval, you have to substitute the x is equal to 0, 1, 2, 3 and so on. Where the sign of f of x changes either from negative to positive or from positive to negative, then the corresponding value of x is the interval. Now, to find the interval here, you have to substitute x is equal to 1, 2, 3 and so on. We observe that f of 1 is equal to minus pi which is less than 0 and f of 2 is equal to minus 3.301 which is also less than 0 that is negative and f of 3 is equal to minus 1.477 which is also less than 0 and the f of 4 is equal to 0.3979 which is greater than 0. Since f of 3 is equal to minus 1.477 which is negative and f of 4 is equal to 0.397 and which is the positive such that f of 3 into f of 4 is less than 0. Therefore, the root lies in the interval 3, 4. Come to an example. Use the regular policy method to find the root of the equation 3x plus sin x is equal to e to the power x correct to 4 decimal of places. Solution, let f of x is equal to 3x plus sin x e to the power x. First we have to find the interval by putting x is equal to 0, 1, 2, 3 and so on whereas sin of f of x changes either from positive to negative or from negative to positive that is the f of 0 is equal to minus 1 which is less than 0 and f of 1 is equal to 1.123 which is greater than 0. Since f of 0 is equal to minus 1 which is less than 0 and f of 1 is equal to 1.123 which is greater than 0. Therefore, the root lies in the interval 0, 1. It is observed that the value of f of x at x is equal to 0 being f of 0 is equal to minus 1 is nearer to 0 as compared to f of 1 is equal to 1.123 and we expect the root in the neighbor of 0. For that we have to find the f of 0, 0.1, f of 0.2, f of 0.3 and so on where the sin of f of x again changes from the positive to negative or from negative to positive that f of 0.3 is equal to minus 0.154 which is less than 0 and f of 0.4 is equal to 0.097 which is greater than 0. Since f of 0.3 is equal to minus 0.154 which is less than 0 and f of 0.4 is equal to 0.097 which is greater than 0. Therefore, the root lies in the interval 0.3 comma 0.4. The regular policy formula is 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. First iteration here the a is equal to 0.3 and b is equal to 0.4 and f of a is equal to minus 0.154 and f of b is equal to 0.097 substituting these values in equation 1 and simplifying we get x1 that is the x1 it gets the first iteration value x1 is equal to 0.3614 substitute the value of x is equal to x1 which is equal to 0.3614 in f of x. It becomes the f of 0.3614 is equal to 0.00244 which is the positive. Since f of 0.3 is equal to minus 0.154 which is less than 0 and f of 0.3614 which is equal to 0.00244 which is greater than 0 such that f of 0.3 into f of 0.3614 is less than 0. Therefore, the root lies in the interval 0.3 comma 0.3614. Second iteration here a is equal to 0.3 and b is equal to 0.3614 and f of a is equal to minus 0.154 and f of b is equal to 0.00244 substituting these values in equation 1 and simplifying we get the x2 is equal to 0.3604 substitute the value of x is equal to x2 which is equal to 0.3604 in f of x it becomes f of 0.3604 is equal to minus 0.0000154. Since f of 0.3604 is equal to minus 0.0000154 is less than 0 and f of 0.3614 which is equal to 0.00244 which is greater than 0 such that f of 0.3604 into f of 0.3614 is less than 0. Therefore, the root lies in the interval 0.3604 comma 0.3614 third iteration here a is equal to 0.3604 and b is equal to 0.3614 and f of a is equal to minus 0.0000154 and f of b is equal to 0.00244 substitute these values in equation 1 and simplifying we get x3 is equal to 0.3604 comparing the x2 and x3 we observe that they are same up to 4 decimal places. Hence the approximate root of the given equation is x is equal to 0.3604 correct to 4 decimal places. Come to another example find the negative root of the equation 2 to the power x minus x minus 3 is equal to 0 using the regla falsi method correct to 4 decimal places. Let f of x is equal to 2 to the power x minus x minus 3 to find the negative root here we have to find the interval we have to put the x is equal to 0 x is equal to minus 1 x is equal to minus 2 x is equal to minus 3 and so on where the sign of f of x changes from negative to positive or from positive to negative then the corresponding value of x are taken as the interval for that we have to put x is equal to 0 that is f of 0 is equal to minus 2 that which is less than 0 and f of minus 1 is equal to minus 1.5 which is also less than 0 and f of minus 2 is equal to minus 0.75 which is less than 0 and f of minus 3 is equal to 0.125 which is greater than 0. Since f of minus 2 is less than 0 and f of minus 3 is greater than 0 such that f of minus 3 into f of minus 2 is less than 0 therefore the root lies in the interval minus 3 minus 2. Now to find the smallest interval it is observed that the value of f of x x is equal to minus 3 is being f of minus 3 is equal to 0.125 is nearer to 0 as compared to f of minus 2 is equal to minus 0.75 and we expect the root in the neighbor of minus 3 that is f of minus 2.9 is equal to 0.039 which is greater than 0 and f of minus 2.8 is equal to minus 0.0564 which is less than 0 since f of minus 2.9 is equal 0.0339 which is greater than 0 and f of minus 2.8 is equal to minus 0.0564 which is less than 0. Such that f of minus 2.9 into f of minus 2.8 is less than 0. Therefore, the root lies in the interval minus 2.9 comma minus 2.8. The regular policy formula is x is equal to a into f of b minus b into f of a whole divided by f of b minus f of b. Call it is equation 1. First iteration here a is equal to minus 2.9 and b is equal to minus 2.8 and f of a is equal to 0.0339 and f of b is equal to minus 0.0564. Substitute these values in equation 1 and simplify we get x1 is equal to minus 2.8625. Substitute the value of x is equal to x1 which is equal to minus 2.8625 in f of x that is f of minus 2.8625 is equal to minus 0.00000333. Since f of minus 2.9 is equal to 0.0339 which is greater than 0 and f of minus 2.865 is equal to minus 0.00000333 which is less than 0 such that f of minus 2.9 into f of minus 2.8625 is less than 0. Therefore, the root lies in the interval minus 2.9 comma minus 2.8625. Second iteration here a is equal to minus 2.9 and the b is equal to minus 2.8625 and the f of a is equal to 0.0339 and f of b is equal to minus 0.00000333. Substitute these values in equation 1 and simplifying we get x2 is equal to minus 2.8625. Comparing the x1 and x2 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 minus 2.8625 correct to 4 decimal of places.