exponential equation has non positive roots 
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots

Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \bigg(-\infty,5-4\sqrt{2}\bigg]\cup \bigg[5+4\sqrt{2},\infty\bigg).$$
but answer is different from that , i did not know where i
am missing. could some help me to solve it. thanks
 A: Hint:
your equation has non positive roots if $0<2^x<1$,  so, with your substitution, the condition becomes $0<y<1$, for the equation $y^2-(a-3)y+(a+4)=0$ 
Can you do from this?

If $y_1,y_2$are the solutions we have :
$
y_1+y_2=a-3 \qquad ,\qquad y_1y_2=a+4
$
and  we want:
$$
\begin{cases}
0<a-3<2\\
0<a+4<1
\end{cases}
$$
 but: $a+4<1 \Rightarrow a<-3$ and $a-3>0 \Rightarrow a>3$
so it is impossible to have $0<y_1<1$ and $0<y_2<1$
A: Rewrite the equation as
$$
a=\frac{4^x+3\cdot 2^x+4}{2^x-1}
$$
and plot the function $a(x)$. From the plot you can see the values of $a$ that correspond to $x<0$.
You can change $y=2^x$ to simplify drawing, but then you will need to neglect negative $y$ as they do not correspond to real values of $x$.
Differentiate to get two critical points $y=1\pm 2\sqrt{2}$ among which only $1+2\sqrt{2}$ is positive, then study intervals of monotonicity and asymptotics. The answer is $a<-4$.

A: You want $f(y) = y^2 - (a-3) y + (a+4)$ to have at least one root in $(0,1]$.
Note that $f(0) = a+4$ and $f(1) = 8$, while the minimum value of $f(y)$, occurring at 
$y = (a-3)/2$, is $(-a^2 + 10 a + 7)/4$.  There are several cases to consider.


*

*If $a < -4$, $f(0) < 0 < f(1)$ so there is a root in $(0,1]$.

*For $a = -4$, $f(y) = y^2 + 7 y$ so the roots $0$ and $-7$ are not in $(0,1]$. 

*For $a > -4$, $f(0) > 0$ and $f(1) > 0$ so the only way to have a root in $(0,1]$ would be if the minimum occurs between $0$ and $1$ and the minimum value $\le 0$.  The
minimum is at $(a-3)/2$, so this requires $a \in (3,5]$.  But for $a$ in this interval,
$-a^2 + 10 a + 7 > 0$.


So the answer is $a < -4$.
Alternative method:
For given $y$, $f(y) = 0$ for $a = A(y) = 4 + y + 8/(y-1)$.  Thus you want the set of values $A(y)$ for $0 < y < 1$ (of course $A(1)$ is undefined).
$A'(y) =  \dfrac{y^2 - y - 7}{(y-1)^2} < 0$ in this interval, so $A$ is a decreasing function of $y$, with $A(y) \to -\infty$ as $y \to 1-$.  Thus the answer is $(-\infty, A(0)) = (-\infty, -4)$.  
A: Not only the equation in $y$ must have real roots, but these roots have to be $\le 1$, since it's asked that the equation in $x$ has non-positive roots, so that $y=2^x\le 2^0=1$.
To test whether the roots are less than $1$, we can place $1$ w.r.t. the roots $y_1,y_2$  of $\;p(y)=y^2-(a-3)y+a+4$. The standard method is to determine the sign of $p(1)=8>0$. Therefore $1$ does not separate the roots of $p(y)$, and  $1$ is either less than or greater than both of them. Furthermore, the half-sum of the roots is $\;\frac{a-3}2$ by Vieta's relations, so that
$$y_1, y_2 \le 1\iff \frac{y_1+y_2}2=\frac{a-3}2 \le 1\iff a\le 5. $$
So the solution is
$$a\in \Bigl\{\Bigl(-\infty,5-4\sqrt{2}\Bigr]\cup \Bigl[5+4\sqrt{2},\infty\Bigr)\Bigr\}\cap (-\infty,5]=\Bigl(-\infty,5-4\sqrt{2}\Bigr].$$
