import java.io.*;
public class quadratic
{
private String str;
public quadratic()
{
str = "";
}
public quadratic(String str)
{
this.str = str;
}
public void accept()throws IOException
{
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the equation in terms of x \nEg: 1x^2+4x+5=0 \nAnd you will receive the possible results of x");
str= br.readLine();
}
public void solve()
{
String A[]=new String [8];
int l=str.length();
int j=0;
char c=' ';
String w="";
for(int i=0;i<7;i++)
{
while(true)
{
c=str.charAt(j);
if(c!='+'&&c!='-'&&c!='=')
w=w+c;
else
{
if(i==4)
A[i]="0"+w;
else
A[i]=w;
i++;
if(i>6)
break;
A[i]=" "+c;
j++;
break;
}
j++;
if(j>=l)
break;
}
w="";
}
System.out.println(A[0]);
System.out.println(A[2]);
System.out.println(A[6]);
int len=A[0].length();
String M=A[0].substring(0,(len-3));
len=A[2].length();
String B=A[2].substring(0,len-1);
int a=Integer.parseInt(M);
int b=Integer.parseInt(B);
int C=Integer.parseInt(A[4]);
if(A[3].equals(" -"))
C=-C;
if(A[1].equals(" -"))
b=-b;
double D=(double)Math.sqrt(b*b-4*a*C);
double ans1=(-b+D)/(2*a);
double ans2=(-b-D)/(2*a);
System.out.println("The values of x are: "+ans1+" and "+ans2);
}
public static void main(String args[])throws IOException
{
quadratic ob = new quadratic();
ob.accept();
ob.solve();
}
}
You don't need a flow chart for that; just use the quadratic formula directly; most programming languages have a square root function or method. You would only need to do this in many small steps if you use Assembly programming. The formulae would be something like this: x1 = (-b + sqrt(b^2 - 4*a*c)) / (2 * a) and x2 = (-b - sqrt(b^2 - 4*a*c)) / (2 * a) where a, b, and c are the coefficients of the quadratic equation in standard form, and x1 and x2 are the solutions you want.
write a vb program to find the magic square
(Uses Square Root Function) PRINT "Ax^2 + Bx + C = 0" INPUT "A = ", A INPUT "B = ", B INPUT "C = ", C D = B * B - 4 * A * C IF D > 0 THEN DS = SQR(D) PRINT "REAL ROOTS:", (-B - D) / (2 * A), (-B + D) / (2 * A) ELSE IF D = 0 THEN PRINT "DUPLICATE ROOT:", (-B) / (2 * A) ELSE DS = SQR(-D) PRINT "COMPLEX CONJUGATE ROOTS:", (-B / (2 * A)); "+/-"; DS / (2 * A); "i" END IF END IF
write a program that reads in the size of the side of square and then pints a hollow square of that size out of asterisks and blanks?
a triangle then a square :)
Write an algorithm to find the root of quadratic equation
The easiest way to write a generic algorithm is to simply use the quadratic formula. If it is a computer program, ask the user for the coefficients a, b, and c of the generic equation ax2 + bx + c = 0, then just replace them in the quadratic formula.
2000X=Y2KoverZzz?
readuse the answer
An example of a quadratic equation is ( ax2 bx c 0 ), where ( a ), ( b ), and ( c ) are constants and ( x ) is the variable.
Write your program and if you are having a problem post it here with a description of the problem you are having. What you are asking is for someone to do your homework for you.
computer scince
ax2 + bx + c
Write the quadratic equation in the form ax2 + bx + c = 0 then the roots (solutions) of the equation are: [-b ± √(b2 - 4*a*c)]/(2*a)
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.
dejene
You don't need a flow chart for that; just use the quadratic formula directly; most programming languages have a square root function or method. You would only need to do this in many small steps if you use Assembly programming. The formulae would be something like this: x1 = (-b + sqrt(b^2 - 4*a*c)) / (2 * a) and x2 = (-b - sqrt(b^2 - 4*a*c)) / (2 * a) where a, b, and c are the coefficients of the quadratic equation in standard form, and x1 and x2 are the solutions you want.