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To solve the equation \( v^2 + 6v - 23 = 0 \) by completing the square, let’s go through each step:

### Step 1: Move the constant term to the other side
Rewrite the equation to separate the variable terms from the constant:
\[
v^2 + 6v = 23
\]

### Step 2: Complete the square
To complete the square, take half of the coefficient of \( v \) (which is 6), square it, and add it to both sides.

1. Half of 6 is 3.
2. Squaring 3 gives \( 3^2 = 9 \).

Add 9 to both sides:
\[
v^2 + 6v + 9 = 23 + 9
\]
\[
v^2 + 6v + 9 = 32
\]

### Step 3: Rewrite the left side as a square
The left side is now a perfect square:
\[
(v + 3)^2 = 32
\]

### Step 4: Take the square root of both sides
Apply the square root to both sides, remembering to consider both the positive and negative roots:
\[
v + 3 = \pm \sqrt{32}
\]
\[
v + 3 = \pm \sqrt{16 \cdot 2} = \pm 4\sqrt{2}
\]

### Step 5: Solve for \( v \)
Subtract 3 from both sides to isolate \( v \):
\[
v = -3 \pm 4\sqrt{2}
\]

### Step 6: Approximate the answers
Now we’ll calculate the decimal approximations of \( v \):
1. \( v = -3 + 4\sqrt{2} \approx -3 + 5.66 = 2.66 \)
2. \( v = -3 - 4\sqrt{2} \approx -3 - 5.66 = -8.66 \)

### Final Answer
So, the solutions are:
\[
v \approx 2.66 \quad \text{or} \quad v \approx -8.66
\]

Would you like further details on any of the steps, or do you have questions?

Here are some related questions for practice:
1. How would you complete the square for \( x^2 + 8x - 15 = 0 \)?
2. What are the advantages of completing the square over using the quadratic formula?
3. How do you determine which number to add to complete the square for any quadratic equation?
4. Can you convert a completed square form back to the original quadratic form?
5. How would you solve \( x^2 - 10x + 25 = 9 \) by completing the square?

**Tip:** Completing the square is also a useful method to derive the quadratic formula.

Solve by completing the square: v^2 + 6v - 23 = 0

Learn how to solve the quadratic equation v^2 + 6v - 23 = 0 using the completing the square method. This step-by-step solution includes key algebraic concepts, such as creating a perfect square and applying the square root property. Suitable for high school students in Grades 9-10.

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