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solve by completing the square.

x^2-8x+4=-3
x^2-8x=____
x^2-8x___=____
(x___ ___)^2 = ___
(x____ ___)= ___ or (x___ ___=___)
x= __ or __

1 answer

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 \( x = 1 \)