Question

Given a cubic $x^3+p x^2+6 x+q=0$ where $p, q \in R$. If one root of the cubic is $(5-i)$ then the value of $(6 p+q)$, is

• A. 4
• B. 5
• C. 6
• D. 9

Answer 1

Answer: (A)

Let $\alpha=5- i \Rightarrow \beta=5+ i$
Let the $3^{\text {rd }}$ root be $\gamma$.
Now $\alpha+\beta+\gamma=10+\gamma=-p$ ...(1)
Also $\alpha \beta=26$
and $ \alpha \beta+\beta \gamma+\gamma \alpha=6 \Rightarrow 26+\gamma(10)=6 \Rightarrow \gamma=-2$
Also $ \alpha \beta \gamma=- q \Rightarrow 26(-2)=- q \Rightarrow q =52$
$\therefore \alpha+\beta-2=-p \Rightarrow p=-8$
Hence $(6 p+q)=-48+52=4$