sqrt

Why do I get this error? I'm trying to solve an equation like this: ax^2 + bx + c Traceback (most recent call last): File "I:/Taller de Programacion I/Clase 5/11.py", line 6, in <module> x1 = (-b + sqrt(b ** 2 - a * c)) / (2 * a) ValueError: math domain error The math module (which I suppose you are using) doesn't support complex numbers. Either use cmath (python2 and python3) or the power operator ** (python3). This should always work, no matter the sign of the discriminant: x1 = (-b + (b ** 2 - 4 * a * c) ** .5) / 2 / a Example: >>> b = 1 >>> a = 2 >>> c = 3 >>> (-b + (b ** 2 - 4 * a * c) **

The following code is throwing undefined symbol error on Linux. $ cat rms.c /* sqrt example */ #include <stdio.h> #include <math.h> int main () { double param, result; param = 1024.0; result = sqrt (param); printf ("sqrt(%lf) = %lf\n", param, result ); return 0; } $ gcc rms.c /tmp/ccaWecFP.o(.text+0x24): In function `main': : undefined reference to `sqrt' collect2: ld returned 1 exit status If I replace argument to sqrt() with (double)16 then program is compiling and executing. Why is this throwing error in first case. This is a linker error. The linker is missing the implementation of sqrt()

Do you know how to do this simple line of code without error using Boost::multiprecison ? boost::multiprecision::cpp_int v, uMax, candidate; //... v += 6 * ceil((sqrt(uMax * uMax - candidate) - v) / 6); Using MSVC there is an error for "sqrt" and it's possible to fix it with: v += 6 * ceil((sqrt(static_cast<boost::multiprecision::cpp_int>(uMax * uMax - candidate)) - v) / 6); Then there is an error for "ceil" and it's possible to fix it with: namespace bmp = boost::multiprecision; typedef bmp::number<bmp::cpp_dec_float<0>> float_bmp; v += 6 * ceil(static_cast<float_bmp>((sqrt(static_cast<bmp:

#include <iostream> #include <cmath> using namespace std; int main() { double a = sqrt(2); cout << a << endl; } hi this is the program to find sqrt of 2 it prints just 1.41421 in the output how to implement it in way such that it will print 200000 digits after decimal point 1.41421..........upto 200 000 digits Is there any approach to print like this? Here is the code for your question which using GNU GMP library. The code will print 1000 digits of sqrt(2), increase the number in the lines with comment to satisfy your request. #include <stdio.h> #include <gmp.h> int main(int argc, char *argv[]

I'm trying to create a complex number from the square root of a negative number using the following code: include Math z = Complex(sqrt(-9)) But it produces this error: Math::DomainError: Numerical argument is out of domain - "sqrt" from kata2.rb:20:in `sqrt' from kata2.rb:20:in `polinomio' from kata2.rb:34 from /home/howarto/.rvm/rubies/ruby-2.0.0-p247/bin/irb:13:in `<main>' How can I build a complex number from the square root of a negative number? The Math.sqrt function can't compute the square root of negative numbers: irb> Math.sqrt(-1) Math::DomainError: Numerical argument is out of

Everyone knows sqrt function from math.h / cmath in C/C++ - it returns square root of its argument. Of course, it has to do it with some error, because not every number can be stored precisely. But am I guaranteed that the result has some precision? For example, 'it's the best approximation of square root that can be represented in the floating point type used or if you calculate square of the result, it will be as close to initial argument as possible using the floating point type given`? Does C/C++ standard have something about it? For C99, there are no specific requirements. But most