Completing the Square Worksheets - Math Monks - Free Printable

To solve the given quadratic equations by completing the square, we will follow a systematic approach for each problem. Let's go through each equation step by step.

---

Problem 1: \( 3x^2 + 24x + 12 = 0 \)

#### Step 1: Divide the entire equation by the coefficient of \( x^2 \) (which is 3).
\[
x^2 + 8x + 4 = 0
\]

#### Step 2: Move the constant term to the right side.
\[
x^2 + 8x = -4
\]

#### Step 3: Complete the square on the left side.
- Take the coefficient of \( x \), which is 8, divide it by 2, and square it: \( \left(\frac{8}{2}\right)^2 = 16 \).
- Add 16 to both sides:
\[
x^2 + 8x + 16 = -4 + 16
\]
\[
x^2 + 8x + 16 = 12
\]

#### Step 4: Write the left side as a perfect square.
\[
(x + 4)^2 = 12
\]

#### Step 5: Solve for \( x \) by taking the square root of both sides.
\[
x + 4 = \pm \sqrt{12}
\]
\[
x + 4 = \pm 2\sqrt{3}
\]

#### Step 6: Isolate \( x \).
\[
x = -4 \pm 2\sqrt{3}
\]

#### Final Answer:
\[
\boxed{x = -4 \pm 2\sqrt{3}}
\]

---

Problem 2: \( (y - 5)^2 = 29 \)

This equation is already in the form of a completed square. Solve directly by taking the square root of both sides.

\[
y - 5 = \pm \sqrt{29}
\]

#### Step 2: Isolate \( y \).
\[
y = 5 \pm \sqrt{29}
\]

#### Final Answer:
\[
\boxed{y = 5 \pm \sqrt{29}}
\]

---

Problem 3: \( a^2 - 6a - 16 = 0 \)

#### Step 1: Move the constant term to the right side.
\[
a^2 - 6a = 16
\]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( a \), which is -6, divide it by 2, and square it: \( \left(\frac{-6}{2}\right)^2 = 9 \).
- Add 9 to both sides:
\[
a^2 - 6a + 9 = 16 + 9
\]
\[
a^2 - 6a + 9 = 25
\]

#### Step 3: Write the left side as a perfect square.
\[
(a - 3)^2 = 25
\]

#### Step 4: Solve for \( a \) by taking the square root of both sides.
\[
a - 3 = \pm \sqrt{25}
\]
\[
a - 3 = \pm 5
\]

#### Step 5: Isolate \( a \).
\[
a = 3 \pm 5
\]

#### Final Answer:
\[
\boxed{a = 8 \text{ or } a = -2}
\]

---

Problem 4: \( g^2 + 4g - 7 = 0 \)

#### Step 1: Move the constant term to the right side.
\[
g^2 + 4g = 7
\]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( g \), which is 4, divide it by 2, and square it: \( \left(\frac{4}{2}\right)^2 = 4 \).
- Add 4 to both sides:
\[
g^2 + 4g + 4 = 7 + 4
\]
\[
g^2 + 4g + 4 = 11
\]

#### Step 3: Write the left side as a perfect square.
\[
(g + 2)^2 = 11
\]

#### Step 4: Solve for \( g \) by taking the square root of both sides.
\[
g + 2 = \pm \sqrt{11}
\]

#### Step 5: Isolate \( g \).
\[
g = -2 \pm \sqrt{11}
\]

#### Final Answer:
\[
\boxed{g = -2 \pm \sqrt{11}}
\]

---

Problem 5: \( d^2 + 14d + 13 = 0 \)

#### Step 1: Move the constant term to the right side.
\[
d^2 + 14d = -13
\]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( d \), which is 14, divide it by 2, and square it: \( \left(\frac{14}{2}\right)^2 = 49 \).
- Add 49 to both sides:
\[
d^2 + 14d + 49 = -13 + 49
\]
\[
d^2 + 14d + 49 = 36
\]

#### Step 3: Write the left side as a perfect square.
\[
(d + 7)^2 = 36
\]

#### Step 4: Solve for \( d \) by taking the square root of both sides.
\[
d + 7 = \pm \sqrt{36}
\]
\[
d + 7 = \pm 6
\]

#### Step 5: Isolate \( d \).
\[
d = -7 \pm 6
\]

#### Final Answer:
\[
\boxed{d = -1 \text{ or } d = -13}
\]

---

Problem 6: \( e^2 + 14e - 15 = 0 \)

#### Step 1: Move the constant term to the right side.
\[
e^2 + 14e = 15
\]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( e \), which is 14, divide it by 2, and square it: \( \left(\frac{14}{2}\right)^2 = 49 \).
- Add 49 to both sides:
\[
e^2 + 14e + 49 = 15 + 49
\]
\[
e^2 + 14e + 49 = 64
\]

#### Step 3: Write the left side as a perfect square.
\[
(e + 7)^2 = 64
\]

#### Step 4: Solve for \( e \) by taking the square root of both sides.
\[
e + 7 = \pm \sqrt{64}
\]
\[
e + 7 = \pm 8
\]

#### Step 5: Isolate \( e \).
\[
e = -7 \pm 8
\]

#### Final Answer:
\[
\boxed{e = 1 \text{ or } e = -15}
\]

---

Problem 7: \( 4x^2 - 8x + 1 = 0 \)

#### Step 1: Divide the entire equation by the coefficient of \( x^2 \) (which is 4).
\[
x^2 - 2x + \frac{1}{4} = 0
\]

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

#### Step 3: Complete the square on the left side.
- Take the coefficient of \( x \), which is -2, divide it by 2, and square it: \( \left(\frac{-2}{2}\right)^2 = 1 \).
- Add 1 to both sides:
\[
x^2 - 2x + 1 = -\frac{1}{4} + 1
\]
\[
x^2 - 2x + 1 = \frac{3}{4}
\]

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

#### Step 5: Solve for \( x \) by taking the square root of both sides.
\[
x - 1 = \pm \sqrt{\frac{3}{4}}
\]
\[
x - 1 = \pm \frac{\sqrt{3}}{2}
\]

#### Step 6: Isolate \( x \).
\[
x = 1 \pm \frac{\sqrt{3}}{2}
\]

#### Final Answer:
\[
\boxed{x = 1 \pm \frac{\sqrt{3}}{2}}
\]

---

Problem 8: \( b^2 + 4b - 6 = 0 \)

#### Step 1: Move the constant term to the right side.
\[
b^2 + 4b = 6
\]

#### Step 2: Complete the square on the left side.
- Take the coefficient of \( b \), which is 4, divide it by 2, and square it: \( \left(\frac{4}{2}\right)^2 = 4 \).
- Add 4 to both sides:
\[
b^2 + 4b + 4 = 6 + 4
\]
\[
b^2 + 4b + 4 = 10
\]

#### Step 3: Write the left side as a perfect square.
\[
(b + 2)^2 = 10
\]

#### Step 4: Solve for \( b \) by taking the square root of both sides.
\[
b + 2 = \pm \sqrt{10}
\]

#### Step 5: Isolate \( b \).
\[
b = -2 \pm \sqrt{10}
\]

#### Final Answer:
\[
\boxed{b = -2 \pm \sqrt{10}}
\]

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Final Answers:

1. \( \boxed{x = -4 \pm 2\sqrt{3}} \)
2. \( \boxed{y = 5 \pm \sqrt{29}} \)
3. \( \boxed{a = 8 \text{ or } a = -2} \)
4. \( \boxed{g = -2 \pm \sqrt{11}} \)
5. \( \boxed{d = -1 \text{ or } d = -13} \)
6. \( \boxed{e = 1 \text{ or } e = -15} \)
7. \( \boxed{x = 1 \pm \frac{\sqrt{3}}{2}} \)
8. \( \boxed{b = -2 \pm \sqrt{10}} \)