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Dim a, b, c, undroot, root, posx, negx, x1, x2 As Single
Private Sub cmdcompute_Click()
a = txta.Text
b = txtb.Text
c = txtc.Text
If txta.Text = "" Then
MsgBox "Please enter the value of A ", 0 + 32
Exit Sub
End If
If txtb.Text = "" Then
MsgBox "Please enter the value of B ", 0 + 32
Exit Sub
End If
If txtc.Text = "" Then
MsgBox "Please enter the value of C ", 0 + 32
Exit Sub
End If
If a = 0 Then
MsgBox "A can't equal 0 ", 0 + 32
Exit Sub
End If
If (b ^ 2) - 4 * a * c < 0 Then
txtx1.Text = ""
txtx2.Text = ""
MsgBox "Imaginary solution : (b^2)-4*a*c < 0", 0 + 48
Exit Sub
End If
'Calcualtions
undroot = (b ^ 2) - 4 * a * c
root = undroot ^ (0.5)
posx = (b * -1) + root
negx = (b * -1) - root
x1 = posx / (2 * a)
x2 = negx / (2 * a)
'Displaying results
txtx1.Text = " " & x1
txtx2.Text = " " & x2
End Sub
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Write a Qbasic program to find the root of a quadratic equation using the formula if the result is less than 0 the program should print the root are immerginary and end the program?
(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 c program to find the roots of a quadratic equation using if else?
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The degree of a quadratic equation?
The Quadratic Formula song: (my grade saver) To the tune of the jack in the box song X equals negative B plus or minus square root of B squared minus 4AC all over 2A :)
Write a program to find the square root of the quadratic equation using flow chart?
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 program to solve a quadratic equation using the quadratic formula in c plus plus?
#include<stdio.h> #include<conio.h> void main() { float a,b,c,z,d,x,y; clrscr(); printf("Enter the value of a,b,c"); scanf("%f %f %f",&a,&b,&c); d=((b*b)-(4*a*c)); z=sqrt(d); x=(-b+z)/(2*a); y=(-b+z)/(2*a); printf("The Quadratic equation is x=%f and y=%f",x,y); getch(); } This answer does not think about imaginary roots. I am a beginner in C programming. There might be flaws in my code. But, it does well. the code is #include<stdio.h> #include<math.h> main() { float a, b, c; float dis, sqdis, real, imag, root1, root2; printf("This program calculates two roots of quadratic equation of the form ax2+bx+c=0.\n"); printf("\n"); printf(" please type the coefficients a, b and c\n"); printf("\n"); printf("a = "); scanf("%f", &a); printf("\n"); printf("b = "); scanf("%f", &b); printf("\n"); printf("c = "); scanf("%f", &c); printf("\n"); dis = (b*b-4*a*c); if(dis < 0) { sqdis = sqrt(-dis); real = -b/(2*a); imag = sqdis/(2*a); printf(" The roots of the quadratic equations are \n x1\t=\t %f + %f i\n x2\t=\t %f - %f i\n", real, imag, real, imag); } else { sqdis = sqrt(dis); root1 = -b/(2*a)+sqdis/(2*a); root2 = -b/(2*a)-sqdis/(2*a); printf("The two roots of the quadratic equations are %f and %f.\n", root1, root2); } system("pause"); }
