<s>[INST] For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution. 
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.

$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
Now for the existance of solution $\displaystyle 2^x+\frac{3}{2^x}\geq 2\sqrt{3}$ Using $\bf{A.M\geq G.M}$
So $\displaystyle a\left(1+\frac{1}{2^x}\right)\geq 2\sqrt{3}\Rightarrow a\geq \sqrt{3}\cdot \frac{2^x}{2^x+1}\geq \sqrt{3}$
But answer given as $a\geq 2,$ please explain me whats wrong with that, Thanks
 [/INST] Put $t = 2^x$. The equation is $t^2-at-a+3 = 0$. This quadratic should have a positive root. Hence $a^2 + 4(a-3) \geq 0$ which gives $(a+6)(a-2) \geq 0$. Thus $a \geq 2$.
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