--- title: "Quadratic equation roots in FORTRAN" subject: "Computer Programming" chapter: 11 key: "semester_1/computer_programming:fortran-quadratic-roots" source: https://www.bctengineer.com/q/semester_1/computer_programming/fortran-quadratic-roots site: BCT Engineering --- # Quadratic equation roots in FORTRAN Computer Programming · Chapter 11: Programming Language: FORTRAN - Asked in: 2071 Chaitra, 8 marks; 2069 Asar, 8 marks (asked 2 times) - Video: [003: PROGRAM IN FORTRAN TO FIND ROOTS OF QUADRATIC EQUATION](https://www.youtube.com/watch?v=CSTp1_DvrS4) — adventure in mathematics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/computer_programming/fortran-quadratic-roots - Whole chapter: https://www.bctengineer.com/md/semester_1/computer_programming/chapter-11 ## Question Write a FORTRAN program to display nature of roots of a quadratic equation. Calculate and display the roots, if they are real and equal. (Solve quadratic equation and display roots in proper format.) ## Short answer Read coefficients a, b, c. If a=0, output 'not quadratic'. Compute discriminant D=b²-4ac. For D>0, real distinct roots (-b±√D)/(2a); D=0, real equal root -b/(2a); D<0, complex roots with real part -b/(2a) and imaginary part √(-D)/(2a). ## Answer **Method.** For a x² + b x + c = 0, the discriminant is D = b² − 4ac. - D > 0: two real, distinct roots (−b ± √D) / 2a. - D = 0: real and equal roots, x = −b / 2a. - D < 0: complex roots, −b/2a ± i √(−D)/2a. - a = 0: not a quadratic equation. ```fortran PROGRAM QUAD REAL A, B, C, D, X1, X2, RP, IP PRINT *, 'ENTER A, B, C' READ *, A, B, C IF (A .EQ. 0.0) THEN PRINT *, 'NOT A QUADRATIC EQUATION' STOP END IF D = B*B - 4.0*A*C IF (D .GT. 0.0) THEN PRINT *, 'ROOTS ARE REAL AND DISTINCT' X1 = (-B + SQRT(D)) / (2.0*A) X2 = (-B - SQRT(D)) / (2.0*A) WRITE(*,10) X1, X2 ELSE IF (D .EQ. 0.0) THEN PRINT *, 'ROOTS ARE REAL AND EQUAL' X1 = -B / (2.0*A) WRITE(*,20) X1 ELSE PRINT *, 'ROOTS ARE COMPLEX' RP = -B / (2.0*A) IP = SQRT(-D) / ABS(2.0*A) WRITE(*,30) RP, IP, RP, IP END IF 10 FORMAT(' X1 = ', F8.3, ' X2 = ', F8.3) 20 FORMAT(' X1 = X2 = ', F8.3) 30 FORMAT(' X = ', F8.3, ' + ', F8.3, 'I AND ', 1 F8.3, ' - ', F8.3, 'I') STOP END ``` **Sample results (worked by hand)** - a, b, c = 1, −5, 6: D = 1, roots `X1 = 3.000 X2 = 2.000`. - a, b, c = 1, −6, 9: D = 0, roots `X1 = X2 = 3.000`. - a, b, c = 1, 2, 5: D = −16, roots −1.000 ± 2.000 i. The format codes `F8.3` print a real in 8 columns with 3 digits after the decimal point. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the nature of roots decided? A. From D = b^2 - 4ac B. From D = b^2 + 4ac C. From D = 2a - b D. From c alone Answer: A. The discriminant D = b^2 - 4ac decides if roots are distinct real, equal real or complex. (Key point: Discriminant) 2. What are the roots when D = 0? A. Two distinct real roots B. Complex roots C. Real and equal roots x = -b/(2a) D. No roots Answer: C. When D = 0 the square-root term vanishes, leaving a repeated root x = -b/(2a). (Key point: Equal roots) 3. Which D gives complex roots, and how is the imaginary part found in the program? A. D > 0; SQRT(D)/(2a) B. D < 0; SQRT(-D)/ABS(2a) C. D = 0; -b/(2a) D. D < 0; SQRT(D) directly Answer: B. A negative D gives complex roots -b/2a +- i sqrt(-D)/2a, so the program takes the square root of -D. (Key point: Complex roots) 4. Why does the program test a = 0 first? A. Because the equation is then not quadratic and 2a would be a division by zero B. To print the answer in F8.3 C. To make D positive D. To speed up sorting Answer: A. If a = 0 the equation is linear and the formula would divide by 2a = 0. 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