C++ :: Quadratic Formula Loss Of Significance And Simplification
Feb 16, 2015
write codes that could solve a quadratic formula, and my codes are like this:
#include <bjarne/std_lib_facilities.h>
int main()
{
cout << "Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c:
";
cout << " a? ";
double a;
cin >> a;
cout << " b? ";
[code]....
Which runs perfectly, but I have 2 questions:
1. How to simplify these code? On the assignment sheet the professor wrote about using void solve_linear(double b, double c); and void solve_ quadratic (double a, double b, double c);which I currently dont understand how these works. He asked us to write a well-encapsulated (as short as possible) program.
2. These are for extra points: the precision problem of floating numbers: professor asked us to find a way to get the precise answer of it, like this: Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c:
a? 1
b? -20000
c? 1.5e-9
Trying to solve the quadratic equation 1*x*x + -20000*x + 1.5e-09 == 0
Using classical formula: Two roots, x = 20000 and x = 0
Using stable formula: Two roots, x = 20000 and x = 7.5e-14
My guess is that
A. Using code like if(x1*x2=a/c) to check if numbers were approximated.
B. Somehow determine the larger one in x1 and x2.
C. Somehow use that larger one to do something
So, I successfully made a program that will perform the quadratic equation on three numbers, imaginary or real. however, i am now trying to simplify the result, as to get rid of the "/2a" on the bottom. Hence the simplify() function. I just started to create the simplification function, and am attempting to divide the imaginary part of the solution as well as the real part of the solution by 2a. Somehow, it gives the error, "error:invalid operands of types 'int' and 'double *' to binary 'operator*'" on lines 105 and 106. I suspect it has to do with the pointers and references that i am passing as parameters. Also, just an aside, I have never actually seen "/=" be used. It can be, right? I know "+=" can be.
#include <iostream> #include <string> #include <cmath> #include <cstdlib>//simplify the answer using namespace std; int count=0; //prototyping double ans_1(double,double,double);
We are suppose to build a program to do the quadratic formula, which isnt really a issue. my issue is i have a if else loop and my if is being ignored.
note: i know that i can use the i number system but we have been asked to not too and to do this instead.
here is my code
#include <stdio.h> #include <math.h> int main() { float a, b, c, d, e, f, g, h, i; /* i would love to figure out a way to do this without so many variables*/
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I need to read repeatedly data from a MPEG2 file to the buffer of 188 bytes and analyse data bit by bit.
I have the problem with correct bytes reading from file. In my code listed below I have two methods for that.
First one is lossing this bytes which in hex_base mode have 0 at the begining, eg: 03, 0F, etc.
The second method based on read function which need to have buffer as a char (lenght > 1 byte). Because of that I receive different values from that from file in some cases.
How can I properly read such file?
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Code: while(1){ nbytes = recvfrom(socket, buffer, MAX_SIZE, 0, (struct sockaddr *) &srv_addr, &addrlen); if (nbytes != -1) { // packet is received
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Code: #include <math.h> #include <stdio.h> int main(void) { float a,b,c,root_1,root_2; printf("Please enter value a from the quadratic equation
[Code] ......
And I keep getting this error:
Code: /tmp/ccgtUIun.o: In function `main': assign345.c:(.text+0xc7): undefined reference to `sqrt' assign345.c:(.text+0xef): undefined reference to `sqrt' collect2: ld returned 1 exit status
/* Check numbers of arguments, and read input */ validInput = (argc = 4);
[Code] ....
Is a section of the code (the first section). And as you can probably guess, it goes on to calculate for a > 0 etc...
I dont really understand what the validinput section is saying? And a, b and c are never defined so Xcode is just saying a,b,c,root1,root2 are uninitialized and I also dont know what means. Do I need to define these values?
Code: int main(int argc, char *argv[]) { int validInput, solution_type; double a, b, c, root_1, root_2, q;
I am creating a program that solves the quadratic equation ax^2 + bx +c.
I have this program almost complete except the output of the equation in the function called display_quadratic. I need the program to display the variables a,b,c in the equation ax^2 + bx + c but I am having 2 problems. My first problem is that I cannot get the right addition and subtraction signs for the equation.
For instance, if the program had the values for a,b,c as 2,2,3
it will display 2x^2 2x 3
How can I get it to display 2x^2 + 2x + 3? or if it was negative like 2x^2 - 2x - 3?
My next question is how do I get to not display the coefficients that are 1?
I had an if-else statement but no matter what I created it would overlap with another statement and print out twice. Here is the code:
This is my code for the quadratic equation. It keeps telling me that my else is illegal since no matching if statement and my else statement is missing a statement
#include <iostream> #include <cmath> #include <string> #include <iomanip> using namespace std; int main() { string Name;
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#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main () { float princ, rate, comp, savings, interest, rates, year;
I need to calculate the area of a triangle using heron's formula. I wrote the code below and i need to modify it so that a user is required to keep entering the three side until he/she decides to stop. If the three sides entered make an invalid triangle, the user is required to re-enter the values until valid triangle is formed. Then the area is displayed.
#include <iostream> #include <cmath> using namespace std; int main(){ double a,b,c=0; double s,A=0;
[Code] .....
the screen output should like this:
Enter three sides of a triangle: 0 1 2 Error! Re-enter three sides of a triangle: -1 1 2 Error! Re-enter three sides of a triangle: 3 4 5 => 6
Continue (y/n)? y
Enter three sides of a triangle: 1 1 2 Error! Re-enter three sides of a triangle: 6 8 10 => 24
Continue (y/n)? n
Done!
Please note that Continue (y/n)? only displayed after a valid triangle is formed. Otherwise, the user needs to re-enter sides until it's valid.
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for(int i=1;i<getSecretShare+1;i++) { //calculateValue=getSecret+(10*p)+(2 * pow(p,2)); int storeThreshold=getThreshold; calculateValue=getSecret+(10*i)+(2 * pow(i,2)); calculateFinalValue=calculateValue % 17 ; cout<<"the calculated value is "<<calculateFinalValue<<endl; }
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I have gotten the substitution of the x value out. However, i am stuck on how do i calculate the values according to the threshold which the user keys in. For example if the user keys in 3.
It will into account the no. of shares to generate which is for example 5 in this case, then it will then take into account the threshold keyed in , which is 3, i will then calculate using the formula using the no. of shares and also threshold. if the threshold is 3 and the number of shares to generate is 5. I should have the following:
the power depends on the threshold which is t-1, 3-1 =2
So I will have to generate two a values based on that and also calculate them according to the power that is dependent on the threshold the first value being to the power of 1 and the 2nd being to the power of 2 if the threshold is 4.
Then I will generate 3 a values and then calculate them :
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Code: #include <stdio.h> #include <math.h> #define PI 3.14159265 double equation(int n); int