Determine all values of $\displaystyle x$ that satisfy the equation.

i. $\displaystyle x^24^{\frac{x}{2}} - 2x2^{x + 1} - 32^x = 0$

$\displaystyle x^22^x - 2x\cdot2^x - 32^x $

$\displaystyle 2^x\left(x^2 - 4x - 3\right)$

$\displaystyle 2^x\left(x + 2\sqrt{7}\right)\left(x - 2\sqrt{7}\right)$

Values are $\displaystyle x = \pm 2\sqrt{7}$

Also, why can't $\displaystyle 4^{\frac{x}{2}}$ be considered as $\displaystyle \left(\sqrt{4}\right)^x = 2^x$?