Solve the equation: $\;$ $2^{2x+1} - 33 \times 2^{x-1} + 4 = 0$
Given equation: $\;\;$ $2^{2x+1} - 33 \times 2^{x-1} + 4 = 0$
i.e. $\;$ $2^{2x} \times 2^1 - 33 \times 2^x \times 2^{-1} + 4 = 0$
i.e. $\;$ $2 \times \left(2^x\right)^2 - \dfrac{33}{2} \times 2^x + 4 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$
Then, in view of equation $(2)$, equation $(1)$ becomes
$2p^2 - \dfrac{33}{2}p + 4 = 0$
i.e. $\;$ $4p^2 - 33p + 8 = 0$
i.e. $\;$ $\left(4p - 1\right) \left(p - 8\right) = 0$
i.e. $\;$ $p = \dfrac{1}{4}$ $\;$ or $\;$ $p = 8$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{1}{4}$, $\;$ $2^x = \dfrac{1}{4}$
i.e. $\;$ $2^x = 2^{-2}$ $\implies$ $x = -2$
when $\;$ $p = 8$, $\;$ $2^x = 8$
i.e. $\;$ $2^x = 2^3$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = -2, \; 3$