Question

given ((x - 7)^2 = 36), select the values of (x).
(x = 13)
(x = 1)
(x = -29)
(x = 42)
complete
given ((x - 1)^2 = 50), select the values of (x).
(x = -49)
(x = 51)
(x = 1 + 5sqrt{2})
(x = 1 - 5sqrt{2})
done

Explanation:

First Problem: Solve \((x - 7)^2 = 36\)

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 \((x - 7)^2 = 36\). Remember that when we take the square root of a square, we get both a positive and a negative solution. So we have:
\[
x - 7=\pm\sqrt{36}
\]
Since \(\sqrt{36} = 6\), this simplifies to:
\[
x - 7=\pm6
\]

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

We now have two separate equations to solve:

• Case 1: \(x - 7 = 6\)

To solve for \(x\), we add 7 to both sides of the equation:
\[
x=6 + 7
\]
\[
x = 13
\]

• Case 2: \(x - 7=-6\)

Again, we add 7 to both sides of the equation:
\[
x=-6 + 7
\]
\[
x = 1
\]

Second Problem: Solve \((x - 1)^2 = 50\)

Step 1: Take square root of both sides

We start by taking the square root of both sides of the equation \((x - 1)^2 = 50\). As before, we will have a positive and a negative solution:
\[
x - 1=\pm\sqrt{50}
\]
Simplify \(\sqrt{50}\). We know that \(50 = 25\times2\), so \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). So the equation becomes:
\[
x - 1=\pm5\sqrt{2}
\]

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

We now have two separate equations to solve:

• Case 1: \(x - 1 = 5\sqrt{2}\)

To solve for \(x\), we add 1 to both sides of the equation:
\[
x=1 + 5\sqrt{2}
\]

• Case 2: \(x - 1=-5\sqrt{2}\)

Again, we add 1 to both sides of the equation:
\[
x=1-5\sqrt{2}
\]

Final Answers:

• For \((x - 7)^2 = 36\), the values of \(x\) are \(x = 13\) and \(x = 1\).
• For \((x - 1)^2 = 50\), the values of \(x\) are \(x = 1 + 5\sqrt{2}\) and \(x = 1-5\sqrt{2}\).

Answer:

Step 1: Take square root of both sides

We start by taking the square root of both sides of the equation \((x - 1)^2 = 50\). As before, we will have a positive and a negative solution:
\[
x - 1=\pm\sqrt{50}
\]
Simplify \(\sqrt{50}\). We know that \(50 = 25\times2\), so \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). So the equation becomes:
\[
x - 1=\pm5\sqrt{2}
\]

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

We now have two separate equations to solve:

• Case 1: \(x - 1 = 5\sqrt{2}\)

To solve for \(x\), we add 1 to both sides of the equation:
\[
x=1 + 5\sqrt{2}
\]

• Case 2: \(x - 1=-5\sqrt{2}\)

Again, we add 1 to both sides of the equation:
\[
x=1-5\sqrt{2}
\]

Final Answers:

• For \((x - 7)^2 = 36\), the values of \(x\) are \(x = 13\) and \(x = 1\).
• For \((x - 1)^2 = 50\), the values of \(x\) are \(x = 1 + 5\sqrt{2}\) and \(x = 1-5\sqrt{2}\).