Question

If one root of the quadratic equation $ax^2+bx+c=0$ is $3-4i$, then $31a+b+c=$

• A. $0$
• B. $2a$
• C. $2b$
• D. $2c$

Hint:

If a polynomial has real coefficients, complex roots always occur in conjugate pairs.

Answer 1

The Correct Option is D

Step 1: Since the coefficients $a, b, c$ are real and one root is $3-4i$, the other root must be its complex conjugate: \[ 3+4i \]
Step 2: Find the sum of roots: \[ (3-4i)+(3+4i)=6 \] But for a quadratic equation, \[ \text{Sum of roots}=-\frac{b}{a} \] So, \[ -\frac{b}{a}=6 \Rightarrow b=-6a \]
Step 3: Find the product of roots: \[ (3-4i)(3+4i)=9+16=25 \] But, \[ \text{Product of roots}=\frac{c}{a} \] So, \[ \frac{c}{a}=25 \Rightarrow c=25a \]
Step 4: Now evaluate $31a+b+c$: \[ 31a+(-6a)+25a=50a \]
Step 5: Since $c=25a$, \[ 50a=2c \]