Question

Solve for x in the given attachment

Answer 1

${( \frac{1}{2} )}^{ {x}^{2} - 2x } < {( \frac{1}{2}) }^{2} \\ {x}^{2} - 2x < 2 \\ {x}^{2} - 2x - 2 < 0$
On solving, we get
$1 - \sqrt{3} < x < 1 + \sqrt{3}$

Answer 2

$( \frac{1}{2} ) ^{ {x}^{2} - 2x} < \frac{1}{4}$
$= > {2}^{ - ( {x}^{2} - 2x) } < {2}^{ - 2}$
$= > 2x - {x}^{2} < - 2$
solving for the quadratic equation,,,,
$= > {x}^{2} - 2x - 2 = 0$
then using the Sridharacharya formula,,,
$x = 1 + \sqrt{3 } \\ or... \\ x = 1 - \sqrt{3}$