<s>[INST] 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 .
 [/INST] 
$x^2 - (c+3)x + 9 = 0.$ 
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
$$\frac{-19}{2} = \frac{-76}{8} < c.$$
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