Sqrt root

[Faisal440]

Write a program to calculate the following equation for any value of x (user input value):

sqrt (pow(x,3)+pow(x,2)-12*x-12)

[Shishykish]

Is this your homework?

You can include the <cmath> header so you can actually calculate the square root. I would start off something like this in terms of pseudo code

Declare a double to variable x and a variable result
Assign a value to x to initialise it like 0.0
Prompt the user using cout for a number they want to put into the formula
Use the cin command to get the number assigned to x
Assign the number you get for the equation to result
Use cout to display the answer for result

[lastchance]

Faisal440 wrote:

Write a program to calculate the following equation for any value of x (user input value):
sqrt (pow(x,3)+pow(x,2)-12*x-12)

Try x = 0

[Faisal440]

Yes this is my homework and i try a lot but can't solve this.
Plz if any one can solve it help me plz.
I have to submit it tomorrow..plz solve it.i will be grateful to you

[Faisal440]

I have to writer it in a program

[Shishykish]

Hi,

I understand it is urgent, but people here don't really like doing other people's homework for them.

#include <iostream>
#include <cmath>

int main()
{
    std::cout << "Hello World!\n";

   /*
     I have given examples for each step I have provided, use them to write your own

      Declare a double to variable x and a variable result (e.g. double y, sum)
      Assign a value to x to initialise it like 0.0 (e.g. y = 0.0)
      Prompt the user using cout for a number they want to put into the formula 
      (example - cout << "Text here")
      Use the cin command to get the number assigned to x
      (cin >> y)
      Assign the number you get for the equation to result
      (sum = 1 + y)
      Use cout to display the answer for result
       (cout << "The answer is = " << y << endl;)
   */

    return 0;
}

Leave yourself plenty of time to do your homework. I am actually a teacher and poor organisation is your enemy I am sure!

[jonnin]

don't overthink it..
sqrt (pow(x,3)+pow(x,2)-12*x-12)
is correct.
or just say
complex<double> f(complex<double> x){return sqrt(x*x*x+x*x-12*x-12);}
if you don't want to deal with complex, you can check that the sqrt is positive and if not print 'imaginary result' or something.

[dutch]

You can use half the multiplications:

x * (x*x + x - 12) - 12

[againtry]

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    // VARIABLES
    double x = 0.0; // USER VALUE
    double y = 0.0; // ANSWER
    
    // PROMPT FOR INPUT FROM USER
    cout << "Please enter a number: " << endl;
    
    // GET VALUE FROM USER
    cin >> x;
    
    // PUT x IN EQUATION AND GET y
    y = sqrt (pow(x,3)+pow(x,2)-12*x-12);
    
    // PRINT ANSWER
    cout << "Answer: " << y << endl;
    
    return 0;
}

Please enter a number: 
10.2
Answer: 32.1068
Program ended with exit code: 0