# Answer to Question #29183 in C++ for amira kausar

Question #29183

Approximating the square root

Implement the sqrt function. The square root of a number, num, can be approximated by repeatedly performing a calculation using the following formula :

nextGuess = (lastGuess + (num / numGuess)) /2

The initial guess can be any positive value (e.g., 1). This value will be the starting vale for lastGuess. If the difference between nextGuess and lastGuess is less than a very small number, such as 0.0001, you can claim that nextGuess is the approximated square root of num. If not, nextGuess becomes the lastGuess and the approximation process continues.

Implement the sqrt function. The square root of a number, num, can be approximated by repeatedly performing a calculation using the following formula :

nextGuess = (lastGuess + (num / numGuess)) /2

The initial guess can be any positive value (e.g., 1). This value will be the starting vale for lastGuess. If the difference between nextGuess and lastGuess is less than a very small number, such as 0.0001, you can claim that nextGuess is the approximated square root of num. If not, nextGuess becomes the lastGuess and the approximation process continues.

Expert's answer

// Square root.cpp : Defines the entry point for the console application.

#include "math.h"

#include "iostream"

using namespace std;

int main()

{

double lastGuess = 1;

double prevGuess = 0;

double eps = 0.000001;

int numGuess = 1;

double num;

cout << "Enter number for evaluation: ";

cin >> num;

while(abs(lastGuess - prevGuess) > eps)

{

double newGuess = (lastGuess + (num / numGuess)) / 2;

prevGuess = lastGuess;

lastGuess = newGuess;

numGuess++;

}

cout << prevGuess * 1000 << endl; // please make sure that multiplying by 1000 is necessary

& // because it wasn't in your algorithm.

}

#include "math.h"

#include "iostream"

using namespace std;

int main()

{

double lastGuess = 1;

double prevGuess = 0;

double eps = 0.000001;

int numGuess = 1;

double num;

cout << "Enter number for evaluation: ";

cin >> num;

while(abs(lastGuess - prevGuess) > eps)

{

double newGuess = (lastGuess + (num / numGuess)) / 2;

prevGuess = lastGuess;

lastGuess = newGuess;

numGuess++;

}

cout << prevGuess * 1000 << endl; // please make sure that multiplying by 1000 is necessary

& // because it wasn't in your algorithm.

}

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