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How could you estimate the square root of 130? - Answers
There are several methods that are based on trial-and-improvement or iteration and a relatively efficient example of this is the Newton-Raphson method. Define the function f(x) = x2 - 130. Your question can then be re-written as finding the solution to f(x) = 0. Let f'(x) = 2x. [This is the derivative of f(x) but you don't need to know or understand that.] Make an initial guess at the solution to f(x) = 0 and call it x0. Then let xn+1 = xn - f(xn)/f'(xn) for n = 0, 1, 2, ... . Thus, each estimate can be used to make the next (better) estimate. This sequence will soon reach a sufficiently accurate estimate. For example, if you start with x0 = 8 which is a pretty poor estimate since 82 = 64 is less than half of 130, x2 is less than 0.2% away from the true answer. Finally remember that there are two square roots: a positive, as well as a negative one. There is another method that resembles long division except that the divisor is augmented at each step but explaining it using this browser is too difficult for me.
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How could you estimate the square root of 130? - Answers
There are several methods that are based on trial-and-improvement or iteration and a relatively efficient example of this is the Newton-Raphson method. Define the function f(x) = x2 - 130. Your question can then be re-written as finding the solution to f(x) = 0. Let f'(x) = 2x. [This is the derivative of f(x) but you don't need to know or understand that.] Make an initial guess at the solution to f(x) = 0 and call it x0. Then let xn+1 = xn - f(xn)/f'(xn) for n = 0, 1, 2, ... . Thus, each estimate can be used to make the next (better) estimate. This sequence will soon reach a sufficiently accurate estimate. For example, if you start with x0 = 8 which is a pretty poor estimate since 82 = 64 is less than half of 130, x2 is less than 0.2% away from the true answer. Finally remember that there are two square roots: a positive, as well as a negative one. There is another method that resembles long division except that the divisor is augmented at each step but explaining it using this browser is too difficult for me.
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How could you estimate the square root of 130? - Answers
There are several methods that are based on trial-and-improvement or iteration and a relatively efficient example of this is the Newton-Raphson method. Define the function f(x) = x2 - 130. Your question can then be re-written as finding the solution to f(x) = 0. Let f'(x) = 2x. [This is the derivative of f(x) but you don't need to know or understand that.] Make an initial guess at the solution to f(x) = 0 and call it x0. Then let xn+1 = xn - f(xn)/f'(xn) for n = 0, 1, 2, ... . Thus, each estimate can be used to make the next (better) estimate. This sequence will soon reach a sufficiently accurate estimate. For example, if you start with x0 = 8 which is a pretty poor estimate since 82 = 64 is less than half of 130, x2 is less than 0.2% away from the true answer. Finally remember that there are two square roots: a positive, as well as a negative one. There is another method that resembles long division except that the divisor is augmented at each step but explaining it using this browser is too difficult for me.
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