Given an open statement determine if their quantification is true The Question

My Work/Question
My book says for part a, iv is true. I disagree. To show an existential statement is false we have to show that for all x that statement is untrue. There are no negative roots for $x^2 - 7x + 10 = 0$ it's roots are $x = 2,5$. So, it is not the case that for some integer x there is a solution for $p(x)$ where $x<0$. Why does my book say it is true?
For part b I have a similar problem. iii says for some positive integer x if $x^2 - 2x - 3 = 0$ then $x<0$. This is true in another universe but in the universe of positive integers we ignore all non positive integers, so shouldn't this be false, because there are no results in our universe of discourse to show it? It also says iv is true which cannot be the case because there are no negative roots to $x^2 - 2x +10 =3$.
Is my book wrong or am I misunderstanding something here? 
 A: I suppose that what you are misunderstanding is the fact that the conditionals are quantified.
For a.iv : the roots of $p(x)$ are : $2,5$. 
Consider $x=3$; we have that $p(3)$ is false and also $3<0$ is; thus, the conditional is true. So, in the domain of positive integers, we have found an $x$ such that $p(x)→(x<0)$ is true, i.e. $\exists x [p(x)→r(x)]$ is true. 
For b.iii : the roots of $q(x)$ are : $−1,3$. 
If we exclude the negative one, there exists a value for $x$ such that $q(x)→(x<0)$ is true ?
For $x=3$ we have that $q(3)$ is true but $x<0$ is not; thus, the conditional is false for $x=3$. 
But for $x=4$, we have that $q(4)$ is false and also $x<0$ is; thus, the conditional is true. So, in the domain of positive integers, we have found an $x$ such that $q(x)→(x<0)$ is true, i.e. $\exists x [q(x)→r(x)]$ is true.   
For c.iii, again we have $-1,3$ has roots of $q(x)$.
Thus, if we consider a universe with only $2,5$, we have that $q(x)$ is always false; but also $x<0$ is. Thus, in this universe, $q(x)→(x<0)$ is true for all $x$ and so also $∃x[q(x)→r(x)]$ is true. 
For c.iv, insted, we have that $p(x)→(x<0)$ is always false, because $2,5$ are the roots of $p(x)$ but they are positive. Thus $∃x[p(x)→r(x)]$ is false.
