Question

Given a cubic $x^3+p{x}^2+6x+q=0$ where $p,q\in R$. If one root of the cubic is $(5-i)$ then the value of $(6p+q)$, 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 $(6p+q)=-48+52=4$