Question

1. solve each equation.

a. ((x + 2)^2 = 9)
b. ((x - 5)^2 = 25)

Explanation:

Part a: Solve \((x + 2)^2 = 9\)

Step 1: Take square root of both sides

To solve for \(x\), we start by taking the square root of both sides of the equation. Remember that when we take the square root of a number, we consider both the positive and negative roots. So, we have:
\[
\sqrt{(x + 2)^2}=\pm\sqrt{9}
\]
Simplifying both sides, the square root of \((x + 2)^2\) is \(|x + 2|\), but since we are taking both positive and negative roots, we can write it as \(x + 2=\pm3\).

Step 2: Solve for \(x\) in two cases

Case 1: When the right - hand side is positive

If \(x + 2 = 3\), we subtract 2 from both sides of the equation to isolate \(x\):
\[
x=3 - 2
\]
\[
x = 1
\]

Case 2: When the right - hand side is negative

If \(x + 2=-3\), we subtract 2 from both sides of the equation:
\[
x=-3 - 2
\]
\[
x=-5
\]

Part b: Solve \((x - 5)^2 = 25\)

Step 1: Take square root of both sides

We take the square root of both sides of the equation \((x - 5)^2 = 25\). Again, we consider both the positive and negative square roots:
\[
\sqrt{(x - 5)^2}=\pm\sqrt{25}
\]
Simplifying, we get \(x - 5=\pm5\).

Step 2: Solve for \(x\) in two cases

Case 1: When the right - hand side is positive

If \(x - 5 = 5\), we add 5 to both sides of the equation:
\[
x=5 + 5
\]
\[
x = 10
\]

Case 2: When the right - hand side is negative

If \(x - 5=-5\), we add 5 to both sides of the equation:
\[
x=-5 + 5
\]
\[
x = 0
\]

Answer:

a. The solutions for \((x + 2)^2 = 9\) are \(x = 1\) and \(x=-5\).
b. The solutions for \((x - 5)^2 = 25\) are \(x = 10\) and \(x = 0\).