Question

1. the solution set to the equation $(x - 1)^2 - 9 = 0$ is

(1) ${0, 9}$
(3) ${-1, 5}$
(2) ${-7, 3}$
(4) ${-2, 4}$

Explanation:

Step1: Isolate the squared term

We start with the equation \((x - 1)^2 - 9 = 0\). First, we add 9 to both sides of the equation to isolate the squared term.

$$ (x - 1)^2 = 9 $$

Step2: Take square roots

Next, we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative root. So we have:

$$ x - 1 = \pm\sqrt{9} $$

Since \(\sqrt{9}=3\), this simplifies to:

$$ x - 1 = \pm 3 $$

Step3: Solve for \(x\) in both cases

We now solve for \(x\) in two separate equations: one with the positive root and one with the negative root.

Case 1: Positive root

$$ x - 1 = 3 $$

Adding 1 to both sides gives:

$$ x = 3 + 1 = 4 $$

Case 2: Negative root

$$ x - 1 = -3 $$

Adding 1 to both sides gives:

$$ x = -3 + 1 = -2 $$

So the solutions are \(x = -2\) and \(x = 4\), which means the solution set is \(\{-2, 4\}\).

Answer:

(4) \(\{-2, 4\}\)