Question

b) \\(|(x - 1)^2 - 4| = 5\\)

Explanation:

Step1: Apply absolute value rule

If $|A|=5$, then $A=5$ or $A=-5$. So:
$(x-1)^2 - 4 = 5$ or $(x-1)^2 - 4 = -5$

Step2: Solve first equation

Isolate the squared term:
$(x-1)^2 = 5 + 4 = 9$
Take square roots:
$x-1 = \pm\sqrt{9} = \pm3$
Solve for $x$:
$x = 1 + 3 = 4$ or $x = 1 - 3 = -2$

Step3: Solve second equation

Isolate the squared term:
$(x-1)^2 = -5 + 4 = -1$
Since squares are non-negative, no real solutions here.

Answer:

$x=-2$ or $x=4$ (solution set: $\{-2, 4\}$)