Thank you for reporting, we will resolve it shortly
Q.
If all real values of $x$ obtained from the equation $4^{x}-(a-3) 2^{x}+a-4=0$ are non-positive, then $a$ belongs to
Complex Numbers and Quadratic Equations
Solution:
Given equation can be written as
$\left(2^{x}\right)^{2}-(a-4) 2^{x}-2^{x}+a-4=0 $
$\Rightarrow \left(2^{x}-1\right)\left(2^{x}-a+4\right)=0$
$\Rightarrow 2^{x}=1,2^{x}=a-4$
Since $x \leq 0$ and $2^{x}=a-4$
$(\because x$ is non-positive)
$\therefore 0< a-4 \leq 1$
i.e., $4< a \leq 5$
i.e., $a \in(4,5]$