How to find the value of $a$ for which $\;\tan^2x + (a+1)\tan x-(a-3)<0$ is true I wanted to know, how can I find the value of $a$ for which the inequality $\tan^2x + (a+1)\tan x-(a-3)<0$ is true for at least one $x\in(0,\pi/2)$.
I don't know how to proceed, any help is appreciated.
 A: To get you started, try letting $(\tan x) = y$, and determine first the values of $a$ and $y = \tan x$ at which the corresponding quadratic equation is equal to zero:
$$\tan^2x + (a+1)\tan x-(a-3)<0\tag{given}$$ 
$$y^2 + (a+1) y - (a - 3) = 0\tag{2}$$
A: Set $\tan x = y$. Note that $y$ can take only positive values since $x\in (0,\dfrac{\pi}{2})$.
Then $$y^2 + (a+1)y - (a-3) <0$$
 $$\iff\left(y+\dfrac{a+1}2 \right)^2 - \dfrac{(a+1)^2}4- (a-3) <0\quad \text{(completing the square)}$$
 $$\iff \left(y+\dfrac{a+1}2 \right)^2 < \dfrac{a^2+6a-11}4 = \dfrac{(a+3)^2-20}4$$
So,
$$ -\sqrt{\dfrac{(a+3)^2-20}4}<y+\dfrac{a+1}2 < \sqrt{\dfrac{(a+3)^2-20}4}$$
$$\iff  -\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2<y < \sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2$$
We just need to make sure that there exists a positive $y$ that satisfies this.
So we need to make sure that the upper bound is positive.
Thus, $\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2>0\iff \sqrt{(a+3)^2-20}>a+1$ 
If $a+1$ negative then the inequality is true, and if $a+1\geq0$, we can square both sides and simplify to obtain $a>3$ which is consistent with $a+1\geq0$. Also, we need $(a+3)^2-20$ to be non-negative, so $(a+3)^2\geq20\iff a\geq\sqrt{20}-3 \text{ or } a\leq-\sqrt{20}-3$.
Thus, 
$ a>3 \text{ or } a\leq-\sqrt{20}-3$.
A: We are really looking at $t^2+(a+1)t-(a-3)=0$, and want at least one positive root.
First observe that if $a\gt 3$ then the constant term is negative, so there is a positive root and a negative root. 
For $a=3$, the larger root $0$ barely misses our interval. 
For $a \lt 3$, the real roots, if any, have the same sign, since their product is positive.
For the rest, we use the fact that the negative of the coefficient of $t$ is the sum of the (possibly non-real) roots. 
As  $a$ decreases from $3$ for a while, the coefficient of $t$ is positive. So the real roots, if any, are both negative. 
Below $a=-1$, the real roots, if any, are positive. This is because the coefficient of $t$ is the negative of the sum of the roots.  
But we need to make sure that the roots are real. For that, the discriminant $(a+1)^2+4(a-3)$ must be $\ge 0$. For negative $a$ this happens at $a=-3-\sqrt{20}$ and below.
Conclusion: There is a positive root if $a\gt 3$ or if $a\le -(3+\sqrt{20})$. 
A: Hint : 
put: $u=\tan x$ and you need the $\Delta >0$ of the quadratic equation $u^2+(a+1)u-(a-3)$=0 by take the numbers $\sqsupset -\infty,c\sqsubset$ and $\sqsupset d,\infty\sqsubset$ where  c and d  is the solution of $0=(a+1)^2+4(a-3)$  and $d>c$
then to be $u^2+(a+1)u-(a-3)<0$ we take the numbers from t to m where t and m is the solution of $u^2+(a+1)u-(a-3)=0$ and $m>t$
