Program to find the square root of a number in c and CPP,c plus plus
In this tutorial, we will try to learn the followings;
- How to draw the flow chart of the square root of a number?
- How to find the square root of a number in C++?
- How to find the square root of a number in C?
Flow chart of the square root of a number
The square root of a number in C++
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#include <iostream> #include<math.h> using namespace std; int main() { float NO, answer_of_GivenNO; cout <<"Enter number to find its squareroot: "; cin >> NO; answer_of_GivenNO = pow(NO, 0.5); cout<<"Squreroor of "<<NO<<" is: " << answer_of_GivenNO << endl; } |
Output:
Square root of only a Positive Number repeatedly using If else statement in C++
Let’s see the “program of the square root of a positive number in C++”.
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#include<iostream> #include<Math.h> using namespace std; int main() { float no; cout<<"Enter a number to find its square root: "; cin>>no; if(no >=0 ) { cout<<"the square root of no is "<<sqrt(no)<<endl; } else { cout<<"Sorry it is a negative number"<<endl; } return 0; } |
Output Sample 1
Enter a number to find its square root: -4
Sorry it is a negative number
Output Sample 2
Enter a number to find its square root: 4
the square root of no is 2
The square root of a number repeatedly using nested if statement in C++
Let’s see the “program of the square root of a number using nested if statement in C++”.
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#include<iostream> #include<Math.h> using namespace std; int main() { float no; cout<<"enter a number to find its squareroot : "<<endl; cin>>no; if(no==0){ cout<<"You Entered 0 So the Square Root Is 0 "<<endl; } else if(no > 0 ) { cout<<"squareroot of no is "<<sqrt(no)<<endl; } else { cout<<"Sorry it is a negative number"<<endl; } return 0; } |
The square root of a number repeatedly using a do-while loop in C++
Let’s see the “program of the square root of a number repeatedly using do-while Loop in C++”.
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#include<iostream> #include<Math.h> using namespace std; int main() { int no; int times; cout<<"Enter No of times you want to Find Square Root "<<endl; cin>>times; int i=0; do{ float no; cout<<"enter a number to find its squareroot : "; cin>>no; if(no >=0 ) { cout<<"squareroot of no is "<<sqrt(no)<<endl; } else { cout<<"complex root"<<endl; } i++; }while(i<times); return 0; } |
Output:
Enter No of how many times you want to Find Square Root
2
Enter Value
3
Enter Value
4
The square root of 3 is 1.73205
The square root of 4 is 2
Square root of a number repeatedly using a For loop in C++
Let’s see the “program of the square root of a number repeatedly using For Loop in C++”.
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#include<iostream> #include<Math.h> using namespace std; int main() { int no; int times; cout<<"Enter No of times you want to Find Square Root "<<endl; cin>>times; for(int i=0;i<times;i++){ float no; cout<<"enter a number to find its squareroot : "; cin>>no; if(no >=0 ) { cout<<"squareroot of no is "<<sqrt(no)<<endl; } else { cout<<"complex root"<<endl; } } return 0; } |
Output:
Enter No of how many times you want to Find Square Root
2
Enter Value
3
Enter Value
4
The square root of 3 is 1.73205
The square root of 4 is 2
The square root of a number repeatedly using Array in C++
Let’s see the “program of the square root of a number using arrays in C++”.
Array: float array[HowManyTimes];
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#include<iostream> #include<Math.h> using namespace std; int main() { int HowManyTimes; cout<<"Enter No of How Many Times you want to Find Square Root "<<endl; cin>>HowManyTimes; float array[HowManyTimes]; for(int i=0;i<HowManyTimes;i++){ cout<<"Enter Value "<<i+1<<endl; cin>>array[i]; } for(int i=0;i<HowManyTimes;i++){ if(array[i] >=0 ) { cout<<"The Square root of "<<array[i]<<"is "<<sqrt(array[i])<<endl; } else { cout<<"complex root"<<endl; } } return 0; } |
Output:
Enter No of how many times you want to Find Square Root
2
Enter Value
3
Enter Value
4
The square root of 3 is 1.73205
The square root of 4 is 2
The square root of a number in C?
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#include <stdio.h> #include <math.h> int main() { double no, root_of_no; printf("Enter number to find its root: "); scanf("%lf", &no); root_of_no = sqrt(no); printf("Square root of %.2lf = %.2lf", no, root_of_no); } |
Output:
