Quadratic Equations: Homework on Non-Real Roots

[Saitama]

Homework Statement

Let a,b,c be real numbers with a>0 such that the quadratic equation $ax^2+bcx+b^3+c^3-4abc=0$ has non real roots. Let $P(x)=ax^2+bx+c$ and $Q(x)=ax^2+cx+b$. Which of the following is true?
a) $P(x)>0 \forall x \in R$ and $Q(x)<0 \forall x \in R$
b) $P(x)<0 \forall x \in R$ and $Q(x)>0 \forall x \in R$
c) neither $P(x)>0 \forall x \in R$ nor $Q(x)>0 \forall x \in R$
d) exactly one of P(x) or Q(x) is positive for all real x.

Homework Equations

The Attempt at a Solution

The first equation has non real roots which its discriminant is less than zero.
$$b^2c^2-4a(b^3+c^3-4abc<0$$
$$\Rightarrow b^2c^2-4ab^3-4ac^3+16a^2bc<0$$
$$\Rightarrow b^2(c^2-4ab)-4ac(c^2-4ab)<0$$
$$\Rightarrow (b^2-4ac)(c^2-4ab)<0$$

$b^2-4ac$ is the discriminant of P(x) and $c^2-4ab$ is the discriminant for Q(x) and both the discriminants are less than which means both P(x) and Q(x) are greater than zero for all $x \in R$.

But there is no option which matches my conclusion.

Any help is appreciated. Thanks!

 

[jbunniii]

Homework Statement

Let a,b,c be real numbers with a>0 such that the quadratic equation $ax^2+bcx+b^3+c^3-4abc=0$ has non real roots. Let $P(x)=ax^2+bx+c$ and $Q(x)=ax^2+cx+b$. Which of the following is true?
a) $P(x)>0 \forall x \in R$ and $Q(x)<0 \forall x \in R$
b) $P(x)<0 \forall x \in R$ and $Q(x)>0 \forall x \in R$
c) neither $P(x)>0 \forall x \in R$ nor $Q(x)>0 \forall x \in R$
d) exactly one of P(x) or Q(x) is positive for all real x.

Homework Equations

The Attempt at a Solution

The first equation has non real roots which its discriminant is less than zero.
$$b^2c^2-4a(b^3+c^3-4abc<0$$
$$\Rightarrow b^2c^2-4ab^3-4ac^3+16a^2bc<0$$
$$\Rightarrow b^2(c^2-4ab)-4ac(c^2-4ab)<0$$
$$\Rightarrow (b^2-4ac)(c^2-4ab)<0$$

$b^2-4ac$ is the discriminant of P(x) and $c^2-4ab$ is the discriminant for Q(x) and both the discriminants are less than which means both P(x) and Q(x) are greater than zero for all $x \in R$.

$(b^2-4ac)(c^2-4ab)<0$ means that one of the discriminants is negative, and the other is positive.

 

[Saitama]

$(b^2-4ac)(c^2-4ab)<0$ means that one of the discriminants is negative, and the other is positive.

Oh yes, thanks!

This means that the answer is c?

 

[jbunniii]

Oh yes, thanks!

This means that the answer is c?

If one of the discriminants is positive, that means the corresponding quadratic has real roots, right? So it can't be c.

 

[Saitama]

If one of the discriminants is positive, that means the corresponding quadratic has real roots, right? So it can't be c.

Woops, I meant d, I switched the options in my mind.

 

[jbunniii]

Woops, I meant d, I switched the options in my mind.

At least it wasn't an exam!