Question

solve by completing the square.
x^2-8x+4=-3
x^2-8x=____
x^2-8x___=____
(x___ ___)^2 = ___
(x____ ___)= ___ or (x___ ___=___)
x= __ or __

Answer 1

To solve the equation \( x^2 - 8x + 4 = -3 \) by completing the square, let's follow the given format.

Starting with:
\[
x^2 - 8x + 4 = -3
\]

First, rearrange the equation:
\[
x^2 - 8x + 4 + 3 = 0
\]
\[
x^2 - 8x + 7 = 0
\]

Now, we need to complete the square for \( x^2 - 8x \):

1. **Starting point:**
\( x^2 - 8x = \_\_\_\_ \)

2. **Take half of the coefficient of \( x \) (which is -8), square it, and add it:**
Half of \(-8\) is \(-4\) and \((-4)^2 = 16\). So, we have:
\[
x^2 - 8x + 16 = \_\_\_\_
\]

3. **Rewrite the equation:**
We can express it as:
\[
(x - 4)^2
\]

4. **Set the equation:**
So now we have:
\[
(x - 4)^2 = 16 - 7
\]
\[
(x - 4)^2 = 9
\]

5. **Square root both sides:**
\[
x - 4 = 3 \quad \text{or} \quad x - 4 = -3
\]

6. **Solve for \( x \):**
- For \( x - 4 = 3\):
\[
x = 3 + 4 = 7
\]
- For \( x - 4 = -3\):
\[
x = -3 + 4 = 1
\]

Thus, the solutions are:
\[
x = 7 \quad \text{or} \quad x = 1
\]

### Final format completed:
- \( x^2 - 8x = 7 \)
- \( x^2 - 8x + 16 = 16 \)
- \( (x - 4)^2 = 9 \)
- \( (x - 4) = 3 \) or \( (x - 4) = -3 \)
- \( x = 7 \) or \( x = 1 \)