If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$ 
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$

I don't know how to find all possible values.
I tried making the lower of the two roots as negative. That does not seem to work out. As $4>0$ the curve will be concave up. There can be numerous cases possible.
 A: Having distinct real roots requires a positive discriminant, i.e.
$$a^2-16(a-3)=(a-4)(a-12)>0,$$
or$$\color{green}{a<4\lor a>12}.$$
Then, having at least one negative root requires that either the sum or the product of the roots be negative, i.e.
$$-a<0\lor a-3<0,$$which is always true and doesn't restrict the solution set !
A: You don't have to solve for the roots (which leads to irrational inequations!) to answer.
Indeed, if $f(x)<0$ for at least one $x$, it has two real roots.The discriminant is $\;\Delta=a^2-16a+48=(a-4)(a-12)$, hence it has two real roots if $a<4$ or if $a>12$.
Suppose $\Delta>0$; we'll determine the sign of the roots: $\;f(0)=a-3$, hence:


*

*if $a<3$, $0$ separates the roots, i.e. there is one negative and one positive root.

*if $a>3$, $0$ is outside of the interval of the roots, i.e. they're both positive or both negative. In that case, their sign is also the sign of their half-sum, which is $-\dfrac a8$. Thus both are negative.

*if $a=3$, the roots are $0$ and $-\dfrac34$.


In all cases, $f(x)$ has at least one negative root, so that $a$ can take any real value in $\,]-\infty,4[\:\cup\:]12,+\infty[$.
A: Your polynomial is $p(x)=4x^2+ax+a-3$  its discriminant is $\Delta=a^2-16a+48$ and its roots are $\frac{-a\pm \sqrt{\Delta}}{8}$. Now there is a $x$ for which value is negative, and it is a parabola opening upwards, tells us that both roots are real and thus $\Delta>0 \implies a^2-16a+48=(a-4)(a-12)>0$ which happens iff $a\in (-\infty,4)\cup(12,\infty)$.
A: The function will be positive when $x$ is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative.
The smallest root is:
$$\frac{-a -\sqrt{a^2-16(a-3)}}{8}$$
So we want to know when $a^2-16(a-3) > 0$, or $a^2-16a+48 > 0$, or $(a-8)^2-16 > 0$, or $(a-8)^2 > 16$, or $$a-8 > 4 \vee a-8 < -4$$
$$a > 12 \vee a < 4$$
When $a$ is positive, both $-a$ and $-a -\sqrt{a^2-16(a-3)}$ are negative, so then the smallest root is negative.
When $a$ is negative, $$\sqrt{a^2-16a+48}=\sqrt{(a-6)^2+12}>\sqrt{(a-6)^2}=|a-6|=6-a$$
With the the second because of the increasing nature of $\sqrt{x}$, the fourth because $a$ and thus $a-6$ negative is. Therefore we have $$\frac{-a -\sqrt{a^2-16(a-3)}}{8}<\frac{-a - (6-a)}{8}=\frac{-6}{8}=-0.75$$
Thus the solution is 
$$a > 12 \vee a < 4$$
A: Examining the extrema of $f(x)=4x^2+ax+a-3$ which is a family of upwards open parabolas:
$$
f'(x) = 8 x + a = 0 \Rightarrow x = -\frac{1}{8}a \\
f''(x) = 8 > 0
$$ 
So we have a local minimum at $x = -(1/8)a$. There the function has the value:
$$
f(-(1/8)a) 
= \frac{1}{16}a^2 - \frac{1}{8}a^2 + a - 3 
= -\frac{1}{16}a^2 + a - 3
$$
The minima regarding different $a$ values lie on the curve
$$
m(x) = -4x^2-8x -3
$$

(Large image version)
From
$$
0 = -\frac{1}{16} a^2 + a - 3 \Rightarrow \\
a^2 = 16 a - 48 \Rightarrow \\
(a - 8)^2 = 64 - 48 = 16 \Rightarrow \\
a = 8 \pm 4
$$
we get that the minimum is non-negative for $a \in [4, 12]$ and there we will find no negative value for $f$ at all.
So we have a chance to find negative $f$ values for negative argument only for $a \in I = \mathbb{R} \setminus [4,12]$, where the minimum occurs for a negative function value. 
For positive $a\in I$ the minimum is at $x_m = -(1/8) a < 0$, so those $a$ are valid.
For negative $a \in I$ for $x = -1/2 < 0$ we have 
$f(-1/2) = 1 - a/2 + a - 3 = a/2 - 2 < 0$, so those $a$ are valid as well. 
This gives $I = \mathbb{R} \setminus [4,12]$ as set of all feasible $a$.
A: The graph is a parabola that opens upward. So there are two cases:


*

*$f(0) < 0$

*f has a minimum at $x_0 < 0$ such that $f(x_0) < 0$
In the first case, $f(0) = a -3 < 0 \text{ gives } a<3$
In the second case, $f'(x) = 8x + a$ which has a minimum at $x_0 = -\frac{a}{8}$ where we require $a > 0$. Then $f(x_0) = \frac{a^2}{16} - \frac{a^2}{8} + a - 3 = \frac{-a^2 + 16a - 48}{16} = g(a)$ which is a parabola in a that opens downward, and has roots $a = \frac{-16 \pm \sqrt{256-192}}{-2} = 8 \pm 4$ so $g(a)$ is negative for all $0 < a \le 4$ and $a > 12$.
Putting everything together:
$ a \in (-\infty,4) \lor a \in (12,+\infty)$
A: You can treat this problem with roots indeed, for as soon as you have two distincts roots $x_1$ and $x_2$, all values between the two roots give a negative $f(x)$.
This you can demonstrate easily. 
So to have distinct roots, you need to have a strictly positive discriminant, that is
$a^2-4a+12>0$
You also want to have a negative root, that is $-a +\sqrt{a^2-4a+12}<0$
A: The given expression is a quadratic in $x$ with the leading coefficient $\geq 0$. Since, it is given to be negative for at least one $x$, the conditions are:


*

*Discriminant $\geq 0$, that is $a^2-4a+12>0$

*At least one root is negative. Solve for the roots and make the smaller one less than $0$. 

