Does the second statement imply the first? 
Let $a, b, c, p, q$ be real numbers. Suppose $\{α, β\}$ are the roots
  of the equation $x^2 + 2px+ q = 0$ and $\{α,\frac{1}{β}\}$ are the
  roots of the equation $ax^2 + 2bx+ c = 0$, where $β \notin \{−1, 0, 1\}$.

STATEMENT-1 : $(p^2 − q)(b^2 − ac) ≥ 0$
STATEMENT 2: $b \neq pa$ or $c \neq qa$

My Attempt:

The second statement means that $α+\dfrac{1}{β} \neq α+β$ or $α\dfrac{1}{β} \neq αβ$, which is true as long as $α\neq0$, since $β \notin \{−1, 0, 1\}$. 
But I have two questions:


*

*Is assuming $\alpha \neq 0$ wrong?

*Does statement $2$ imply statement $1$ ?
 A: Statement 1
Note that you can write a second order equation with its roots as:
$$x^2+2px+q=(x-\alpha)(x-\beta)$$
so that $2p=-\alpha-\beta$ and $q=\alpha\beta$. Doing the same with the second equation:
$$a x^2+2bx+c=a(x-\alpha)(x-\frac{1}{\beta})$$
yields $c=a\frac{\alpha}{\beta}$ and $2b=a(-\alpha-\frac{1}{\beta})$. Direct substitution (can you fill in the details?) results in:
$$p^2-q=\frac{1}{4}(\alpha-\beta)^2\ge0$$
and
$$b^2-ac=\frac{a^2}{4}(\alpha-\frac{1}{\beta})^2\ge 0$$
Statement 2
Assume for the sake of contradiction that $b=pa$. Let us first see that also $c=qa$ holds. The second equation becomes:
$$ax^2+2bx+c=ax^2+2pax+c=0$$
And the first equation, multiplied by $a$ is:
$$ax^2+2pax+aq=0$$
so substracting both equations gives $c-aq=0$. 
Now, substituting further $c=aq$ in the second equation we obtain:
$$ax^2+2bx+c=ax^2+2pax+aq=a(x^2+2px+q)=0$$
which is a multiple of the first equation, so both equations have the same roots, i.e. $\beta=\frac{1}{\beta}$ or $\beta=\pm 1$. This is a contradiction, so $b\neq pa$.
A: This is a reply to the comment  

"Does statement 2 imply statement 1?"  

I don't have enough rep to comment.
The answer is yes.
Statement 1 was derived from the condition that $\beta \notin \{-1,0,1\}.$
Since statement 2 also requires that condition on $\beta$, then having that condition present implies both of them, so we can say that if statement 2 is present, statement 1 must also be present.
(For $\alpha \neq 0$ of course.)
