Completing the Square Worksheet Pack - Free Printable

The task involves solving quadratic equations by completing the square. Let's go through each problem step by step and explain the process.

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Problem 1: Solve \( x^2 + 4x + 2 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ x^2 + 4x = -2 \]

#### Step 2: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is 4. Half of 4 is 2, and squaring it gives \( 2^2 = 4 \).
\[ x^2 + 4x + 4 = -2 + 4 \]
\[ x^2 + 4x + 4 = 2 \]

#### Step 3: Write the left side as a perfect square.
\[ (x + 2)^2 = 2 \]

#### Step 4: Take the square root of both sides.
\[ x + 2 = \pm \sqrt{2} \]

#### Step 5: Solve for \( x \).
\[ x = -2 \pm \sqrt{2} \]

So, the solutions are:
\[ x = -2 + \sqrt{2} \quad \text{and} \quad x = -2 - \sqrt{2} \]

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Problem 2: Solve \( 2x^2 - 4x - 36 = 0 \)

#### Step 1: Divide the entire equation by 2 to simplify.
\[ x^2 - 2x - 18 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ x^2 - 2x = 18 \]

#### Step 3: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is -2. Half of -2 is -1, and squaring it gives \( (-1)^2 = 1 \).
\[ x^2 - 2x + 1 = 18 + 1 \]
\[ x^2 - 2x + 1 = 19 \]

#### Step 4: Write the left side as a perfect square.
\[ (x - 1)^2 = 19 \]

#### Step 5: Take the square root of both sides.
\[ x - 1 = \pm \sqrt{19} \]

#### Step 6: Solve for \( x \).
\[ x = 1 \pm \sqrt{19} \]

So, the solutions are:
\[ x = 1 + \sqrt{19} \quad \text{and} \quad x = 1 - \sqrt{19} \]

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Problem 3: Solve \( x^2 + 7x + 12 = 0 \)

#### Step 1: Move the constant term to the right side.
\[ x^2 + 7x = -12 \]

#### Step 2: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is 7. Half of 7 is \( \frac{7}{2} \), and squaring it gives \( \left( \frac{7}{2} \right)^2 = \frac{49}{4} \).
\[ x^2 + 7x + \frac{49}{4} = -12 + \frac{49}{4} \]

Convert -12 to a fraction with a denominator of 4:
\[ -12 = -\frac{48}{4} \]
\[ x^2 + 7x + \frac{49}{4} = -\frac{48}{4} + \frac{49}{4} \]
\[ x^2 + 7x + \frac{49}{4} = \frac{1}{4} \]

#### Step 3: Write the left side as a perfect square.
\[ \left( x + \frac{7}{2} \right)^2 = \frac{1}{4} \]

#### Step 4: Take the square root of both sides.
\[ x + \frac{7}{2} = \pm \frac{1}{2} \]

#### Step 5: Solve for \( x \).
\[ x = -\frac{7}{2} \pm \frac{1}{2} \]

This gives two solutions:
\[ x = -\frac{7}{2} + \frac{1}{2} = -\frac{6}{2} = -3 \]
\[ x = -\frac{7}{2} - \frac{1}{2} = -\frac{8}{2} = -4 \]

So, the solutions are:
\[ x = -3 \quad \text{and} \quad x = -4 \]

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Problem 4: Solve \( x^2 - 2x + 1 = 4 \)

#### Step 1: Simplify the equation.
\[ x^2 - 2x + 1 - 4 = 0 \]
\[ x^2 - 2x - 3 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ x^2 - 2x = 3 \]

#### Step 3: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is -2. Half of -2 is -1, and squaring it gives \( (-1)^2 = 1 \).
\[ x^2 - 2x + 1 = 3 + 1 \]
\[ x^2 - 2x + 1 = 4 \]

#### Step 4: Write the left side as a perfect square.
\[ (x - 1)^2 = 4 \]

#### Step 5: Take the square root of both sides.
\[ x - 1 = \pm 2 \]

#### Step 6: Solve for \( x \).
\[ x = 1 \pm 2 \]

This gives two solutions:
\[ x = 1 + 2 = 3 \]
\[ x = 1 - 2 = -1 \]

So, the solutions are:
\[ x = 3 \quad \text{and} \quad x = -1 \]

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Problem 5: Solve \( 3x^2 + 9x + 6 = 0 \)

#### Step 1: Divide the entire equation by 3 to simplify.
\[ x^2 + 3x + 2 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ x^2 + 3x = -2 \]

#### Step 3: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is 3. Half of 3 is \( \frac{3}{2} \), and squaring it gives \( \left( \frac{3}{2} \right)^2 = \frac{9}{4} \).
\[ x^2 + 3x + \frac{9}{4} = -2 + \frac{9}{4} \]

Convert -2 to a fraction with a denominator of 4:
\[ -2 = -\frac{8}{4} \]
\[ x^2 + 3x + \frac{9}{4} = -\frac{8}{4} + \frac{9}{4} \]
\[ x^2 + 3x + \frac{9}{4} = \frac{1}{4} \]

#### Step 4: Write the left side as a perfect square.
\[ \left( x + \frac{3}{2} \right)^2 = \frac{1}{4} \]

#### Step 5: Take the square root of both sides.
\[ x + \frac{3}{2} = \pm \frac{1}{2} \]

#### Step 6: Solve for \( x \).
\[ x = -\frac{3}{2} \pm \frac{1}{2} \]

This gives two solutions:
\[ x = -\frac{3}{2} + \frac{1}{2} = -\frac{2}{2} = -1 \]
\[ x = -\frac{3}{2} - \frac{1}{2} = -\frac{4}{2} = -2 \]

So, the solutions are:
\[ x = -1 \quad \text{and} \quad x = -2 \]

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Problem 6: Solve \( 2x^2 - 2x + 32 = 0 \)

#### Step 1: Divide the entire equation by 2 to simplify.
\[ x^2 - x + 16 = 0 \]

#### Step 2: Move the constant term to the right side.
\[ x^2 - x = -16 \]

#### Step 3: Add the square of half the coefficient of \( x \) to both sides.
The coefficient of \( x \) is -1. Half of -1 is \( -\frac{1}{2} \), and squaring it gives \( \left( -\frac{1}{2} \right)^2 = \frac{1}{4} \).
\[ x^2 - x + \frac{1}{4} = -16 + \frac{1}{4} \]

Convert -16 to a fraction with a denominator of 4:
\[ -16 = -\frac{64}{4} \]
\[ x^2 - x + \frac{1}{4} = -\frac{64}{4} + \frac{1}{4} \]
\[ x^2 - x + \frac{1}{4} = -\frac{63}{4} \]

#### Step 4: Write the left side as a perfect square.
\[ \left( x - \frac{1}{2} \right)^2 = -\frac{63}{4} \]

#### Step 5: Take the square root of both sides.
Since the right side is negative, there are no real solutions.

So, the solution is:
\[ \text{No real solutions} \]

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Final Answers:

1. \( x = -2 + \sqrt{2}, \, x = -2 - \sqrt{2} \)
2. \( x = 1 + \sqrt{19}, \, x = 1 - \sqrt{19} \)
3. \( x = -3, \, x = -4 \)
4. \( x = 3, \, x = -1 \)
5. \( x = -1, \, x = -2 \)
6. No real solutions

\[
\boxed{x = -2 + \sqrt{2}, \, x = -2 - \sqrt{2}, \, x = 1 + \sqrt{19}, \, x = 1 - \sqrt{19}, \, x = -3, \, x = -4, \, x = 3, \, x = -1, \, x = -1, \, x = -2, \, \text{No real solutions}}
\]