Question

Let $a,b,c\in R$ with $a>0$ such that the equation $a{x}^2+bcx+b^3+c^3-4abc=0$ has non-real roots. Let $P(x)=a{x}^2+bx+c$ and $Q(x)=a{x}^2+cx+b$, then

• A. $P(x)>0$ for all $x\in R$ and $Q(x)<0$ for all $x\in R$.
• B. $P(x)<0$ for all $x\in R$ and $Q(x)>0$ for all $x\in R$.
• C. neither $P(x)>0$ for all $x\in R$ nor $Q(x)>0$ for all $x\in R$.
• D. exactly one of $P(x)$ or $Q(x)$ is positive for all real $x$.

Answer 1

Answer: (D)

$D<0$
$\Rightarrow b^2{c}^2-4\left(b^3+c^3-4abc\right)a<0$
$\Rightarrow \left(b^2{c}^2-4a{b}^3\right)+\left(16a^{2}bc-4a{c}^3\right)<0$
$\Rightarrow b^2\left(c^2-4ab\right)-4ac\left(c^2-4ab\right)<0$
$\Rightarrow \left(b^2-4ac\right)\left(c^2-4ab\right)<0$
$\Rightarrow D_{P(x)}\cdot D_{Q(x)}<0$
$\therefore$ Exactly one of $P(x)$ or $Q(x)$ is positive for all real $x$.