Multiple choice

If $3+4i$ is a root of the equation $a{x}^{2}+bx+c=0$ where $a,b,c\in R$ then $31a+b+c=$

  1. $0$
  2. $2a$
  3. $2b$
  4. $2c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the coefficients are real, the complex conjugate 3-4i must also be a root. The sum of roots is 6 = -b/a and the product is (3+4i)(3-4i) = 9+16 = 25 = c/a. Substituting b = -6a and c = 25a into 31a+b+c gives 31a - 6a + 25a = 50a, which equals 2c (since 2c = 50a).

Keep practicing — related questions