1. Solve the equation $|x - 1| - 3 = 0$. - Start by isolating the absolute value: $|x - 1| = 3$. - Recall that $|a| = b$ means $a = b$ or $a = -b$. - So, $x - 1 = 3$ or $x - 1 = -3$. - Solve each: - $x = 4$ - $x = -2$ 2. Solve the equation $x^{-4} - 13x^{-2} + 36 = 0$. - Let $y = x^{-2}$, so the equation becomes $y^2 - 13y + 36 = 0$. - Factor or use quadratic formula: - Factors of 36 that sum to 13 are 9 and 4. - So, $(y - 9)(y - 4) = 0$. - Thus, $y = 9$ or $y = 4$. - Recall $y = x^{-2} = \frac{1}{x^2}$, so: - $\frac{1}{x^2} = 9 \Rightarrow x^2 = \frac{1}{9} \Rightarrow x = \pm \frac{1}{3}$. - $\frac{1}{x^2} = 4 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$. Final answers: - For problem 15: $x = -2$ or $x = 4$, so option C (Both A and B). - For problem 16: $x = \pm \frac{1}{3}$ and $x = \pm \frac{1}{2}$, so option D.