Given a number, n, and an approximation for its square root, a closer approximat

Question

Given a number, n, and an approximation for its square root, a closer approximation to the actual square root can be obtained by using this formula:

new approximation=((n/ previous approximation)+ previous approximation)/2

Using the information, write a C++ program with Visual Studio 2015 that promts the user for a number and an initial guess as its square root. Using this input data, your program should be able to calculate to 0.00001. (HINT: Stop the loop when the difference between the two approximations is less than 0.00001.)

Explanation / Answer

#include<bits/stdc++.h>
using namespace std;
float squareRootOfNumber(float n)//function to find square
{
  
float a = n;
float y = 1;
float accuracy = 0.00001;
while(a - y > accuracy)
{
a = (a + y)/2;
y = n/a;
}
return a;
}

int main()
{
int n;
cout<<"enter number ";
cin>>n;
cout<<"Square root of "<<n<<" is "<<squareRootOfNumber(n);

}

===================================================

Output:

akshay@akshay-Inspiron-3537:~/Chegg$ g++ squareroot.cpp
akshay@akshay-Inspiron-3537:~/Chegg$ ./a.out
enter number
26
Square root of 26 is 5.09902

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.