Quadratic equation question i have a multiple choice question
Let $x_1$ and $x_2$ be the real root of The quadratic equation $x^2-(c+3)x+9=0$ if $x_1<-2$ and $x_2<-2$, then ......
(a) $c<\frac{-19}{2}$ or $c>9$
(b) $\frac{-19}{2}<c\le-9$
(c) $\frac{-19}{2}<c\le-7$
(d) $-9<c<3$
(e) $c>3$
and the correct answer is (b) , here is the solution, we know that based from the quadratic equations,
$x^2-(c+3)x+9=0$ $\dots(L_1)$
$x_1+x_2=c+3$ and $x_1 x_2=9$
then we know that $x_1<-2$ and $x_2<-2$ this implies
$x_1+2 < 0$ and $x_2+2<0$
then when we add it we obtain ,
$x_1+x_2+4<0$
$c+3+4<0$ , then we get
$c<-7$ $\dots (H_1)$
when we multiply it we obtain ,
$(x_1+2)(x_2+2)>0$
$x_1x_2+2(x_1+x_2)+4>0$
$9+2(c+3)+4>0$ then we will obtain
$c>\frac{-19}{2}$ $\dots (H_2)$
Now, since we are given the roots must be $x_1<-2$ and $x_2<-2$ of $(L_1)$ this means it can be either have a twin roots or two distinct root so this means the discriminant must be larger or equal to zero
$D\geq0$
$(-(c+3))^2-4.1.9 \geq 0$
$c^2+6c+9-36 \geq 0$
$c^2+6c-27 \geq 0$
$(c-3)(c+9) \geq 0$
so to satisfy the equation , $c\leq-9$ or $c\geq3$ $\dots(H_3)$
So the value of $c$ must be the intersection of $(H_!)$,$(H_2)$,$(H_3)$ which is $\frac{-19}{2}<c\le-9$
so the answer is (b)
However, i am still not clear about the question and the logic order of this solution , this question seems like asking what is the value of c so that the value of $x_1$ and $x_2$ will be always less than-2.However the question says "if $x_1<-2$ and $x_2<-2$" that means , that we are given $x_1<-2$ and $x_2<-2$ then it will give a lot of implication of $c$ . This question seems like we do the converse proof , we finding the value of c then it imply the roots are less than 2. i am really confuse about it . the second one is , we are given this condition $x_1+2<0$ and $x_2+2<0$ . then we try to solving the value of c by adding it and multiply it , my question is ,  how does this step ("by adding it and multiply it") can guarantee the value of c so that the value of $x_1$ and $x_2$ will be always less than -2 ? i dont get it . When i analyze the solution and graph seems like they asking what condition of c so that the roots are always less than -2 .
 A: 1st question - You're asking why we find a value of c first, then we take discriminants right? It feels like extra working to you?
The first part of the proof is basically coming up with a value that satisfies the first few conditions. In this instance, you found that $\frac{-19}{2}$ < c < -7.
Even though this gives a value of c, it only gives values of c that satisfies the conditions above. It does not take into consideration whether roots exist when c is in the range you've found. For example, c = -8 is within the range of -19/2 < c < -7, but the graph looks like this
As you can see, this graph has no roots, which is nonsense because obviously the real roots exist. Therefore, we need the discriminants to make sure that the c values that we have give a graph with real roots. Doing discriminants, we get a second range that we can use to find a value of c that
(1) Meets the required conditions
(2) Has real roots that exist.
qn 2 - Asking about why we add and multiply
Well, you have the values for x1 + x2 and x1x2 right? And these are the only 2 values you have, so you need to use them to find the value of c. The only way to make x1x2 is to multiply them together and x1 + x2 is found by adding them together.
A: 
$x^2 - (c+3)x + 9 = 0.$ 
The roots are $x_1, x_2,$ which may or may not be distinct. 
$x_1 < -2, x_2 < -2.$

As I interpret the question, it is asking for all values of $c$ that will satisfy the above constraints.
I would have worded the solution differently.
First of all, you know that 
$x_1 + x_2 = (c + 3) \implies (c + 3) < -4 \implies c < -1.$
Second of all, you know that the two roots are real, which implies that 
$(c + 3)^2 \geq (4 \times 9) = 6^2 \implies c \leq -9,$ 
since it has been established that $c$ must be negative.
So, the problem is reduced to determining for which values of $c$ that are $\leq -9,$ will the two roots, $x_1$ and $x_2$ both be $< -2.$

To determine this, the quadratic formula must be applied.
Then,
$$x_1, x_2 = \frac{1}{2} \left[(c + 3) \pm \sqrt{(c+3)^2 - 36}\right]$$
$$= \frac{1}{2} \left[(c + 3) \pm \sqrt{c^2 + 6c - 27}\right]. \tag1 $$
At this point, since $c \leq -9,$ it is obvious that the root represented by
$$\frac{1}{2} \left[(c + 3) - \sqrt{c^2 + 6c - 27}\right]$$
will definitely be $< -2.$
Consequently, the entire problem has been reduced to determining for which values of $c \leq -9$, will the following root also be $< -2$:
$$\frac{1}{2} \left[(c + 3) + \sqrt{c^2 + 6c - 27}\right].  \tag2 $$

To attack (2) above, I first clear the $~\displaystyle \frac{1}{2}$ fraction, and isolate the expression under the radical.
Thus,
$$(c + 3) + \sqrt{c^2 + 6c - 27} < -4 \implies   $$
$$\sqrt{c^2 + 6c - 27} < -4 - (c + 3) = -(7+c).   \tag3 $$
Examining (3) above, I note that under the constraint that $c \leq -9,$ the LHS and RHS are both non-negative numbers.  Further, I know that (in general), if $0 \leq r,s$ then $r < s \iff r^2 < s^2$.  So, I can attack (3) above by squaring both sides.
This implies that I am looking for the values of $c \leq -9$ such that
$$c^2 + 6c - 27 < c^2 + 14c + 49 \iff -76 < 8c.$$
This simplies to a lower bound for $c$ of
$$\frac{-19}{2} = \frac{-76}{8} < c.$$
