Complete the Square

Grade 9 · algebra · 77 practice problems · read aloud

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Complete the Square: Your Algebra Superpower

What is it? Completing the square is a technique to rewrite a quadratic expression (like x² + 6x + 5) into a perfect square trinomial plus or minus a constant. It's super useful for solving quadratic equations, graphing parabolas easily by finding the vertex, and even deriving the Quadratic Formula! 🧠

Step-by-Step Guide

  1. Start with the quadratic. Ensure the coefficient of x² is 1. If it's not, divide every term by that coefficient.
  2. Move the constant. Rewrite the equation so the x² and x terms are together, and the constant is on the other side of the equals sign.
  3. Find the "magic number." Take the coefficient of the x-term, divide it by 2, and square the result.
  4. Add it to both sides. Add this magic number to both sides of the equation. This makes the left side a perfect square trinomial.
  5. Factor and solve. Factor the perfect square trinomial. You'll get something like (x + a)². Then, take the square root of both sides and solve for x.

Visual Examples

Example 1: Solve x² + 6x + 5 = 0

  1. The x² coefficient is already 1.
  2. Move constant: x² + 6x = -5
  3. Magic number: (6/2)² = (3)² = 9
  4. Add to both sides: x² + 6x + 9 = -5 + 9 → x² + 6x + 9 = 4
  5. Factor & Solve: (x + 3)² = 4 → x + 3 = ±2 → x = -1 or x = -5

Example 2: Solve x² - 4x - 7 = 0

  1. x² coefficient is 1.
  2. Move constant: x² - 4x = 7
  3. Magic number: (-4/2)² = (-2)² = 4
  4. Add to both sides: x² - 4x + 4 = 7 + 4 → x² - 4x + 4 = 11
  5. Factor & Solve: (x - 2)² = 11 → x - 2 = ±√11 → x = 2 ± √11

Common Mistakes to Avoid

⚠️ Forgetting to add the constant to BOTH sides. If you only add it to one side, your equation becomes unbalanced.

⚠️ Using the wrong "magic number." Remember the pattern: (b/2)². Don't forget to divide by 2 first, then square.

⚠️ Sign errors in the binomial. The sign inside (x ± ?)² comes from the sign of the original x-term's coefficient.

Tips & Tricks

Memory Aid: The number you add is always half the middle, squared. Say it out loud!

Vertex Form: The final form (x - h)² + k instantly gives you the parabola's vertex at (h, k). This is a huge graphing shortcut!

Check Your Work: After factoring, expand (x + a)² to make sure you get back your original x² and x terms.

How to Practice

  • Start with simple equations where the x² coefficient is 1 and the x-term coefficient is even.
  • Practice writing the vertex form from standard form without solving the whole equation.
  • Use online quizzes that give instant feedback. Try a mix of solving and graphing problems.

Practice problems

6 of the 77, worked through step by step — try them before opening the answer.

1 x² + 6x + 8 = 0

Hint: Transform the quadratic equation by adding and subtracting a specific value to create a perfect square trinomial, then solve for the variable.

Show the answer

Answer: -2, -4

  1. Identify the coefficients. The equation is in the form ax² + bx + c = 0. Here, a = 1, b = 6, c = 8.
  2. Factor the quadratic expression. We look for two numbers that multiply to a*c = 1*8 = 8 and add to b = 6. The pairs of factors of 8 are: (1,8), (2,4), (4,2), (8,1), and also negative pairs if needed. Check 2 and 4: 2*4 = 8, and 2+4 = 6. Yes, these work.
  3. Write the factored form. Since a = 1, we can factor directly as: (x + 2)(x + 4) = 0.
  4. Apply the zero-product property. If (x + 2)(x + 4) = 0, then at least one factor must be zero. So, either x + 2 = 0 or x + 4 = 0.
  5. Solve each equation. From x + 2 = 0, we get x = -2. From x + 4 = 0, we get x = -4.
  6. State the solution. The solutions are x = -2 and x = -4. Final answer: -2, -4

We are solving the quadratic equation: x² + 6x + 8 = 0.

2 x² + 6x + 5 = 0

Hint: Transform the equation so that one side becomes a perfect square trinomial, then solve for the variable.

Show the answer

Answer: x = -1, -5

  1. Identify the coefficients. The equation is in the form ax² + bx + c = 0. Here, a = 1, b = 6, c = 5.
  2. Factor the quadratic expression. We look for two numbers that multiply to a*c = 1*5 = 5 and add to b = 6. The numbers 1 and 5 multiply to 5, but 1 + 5 = 6 — yes, that works.
  3. Write the factored form. Since a = 1, we can factor directly: x² + 6x + 5 = (x + 1)(x + 5) = 0.
  4. Apply the zero-product property. If (x + 1)(x + 5) = 0, then at least one factor must be zero. So: x + 1 = 0 or x + 5 = 0
  5. Solve each equation. x + 1 = 0 → x = -1 x + 5 = 0 → x = -5
  6. State the solution. The solutions are x = -1 and x = -5.

We are solving the quadratic equation: x² + 6x + 5 = 0

3 x² - 4x - 5 = 0

Hint: Transform the quadratic equation into a perfect square form by adding and subtracting the appropriate constant

Show the answer

Answer: x = 5, -1

  1. Start with x² - 4x - 5 = 0
  2. Move constant term to right side: x² - 4x = 5
  3. Find the number to complete the square: take half of -4, which is -2, then square it to get 4
  4. Add 4 to both sides: x² - 4x + 4 = 5 + 4
  5. Write left side as perfect square: (x - 2)² = 9
  6. Take square root of both sides: x - 2 = ±3
  7. Solve for x: x = 2 + 3 = 5 or x = 2 - 3 = -1 The solutions are x = 5 and x = -1.

4 x² + 8x + 13 = ?

Hint: To complete the square, take half of the coefficient of the x-term, square it, and adjust the constant term accordingly.

Show the answer

Answer: (x + 4)² - 3

  1. Identify the goal** We want to write this expression in the form (x + a)² + b, because that matches the given correct answer format. --- **
  2. Complete the square** Take the x² + 8x part: - The coefficient of x is 8. - Half of 8 is 4. - Square 4 to get 16. So: x² + 8x = (x + 4)² - 16 Why? Because (x + 4)² = x² + 8x + 16, so subtracting 16 gives back x² + 8x. --- **
  3. Substitute back into the original expression** x² + 8x + 13 = [ (x + 4)² - 16 ] + 13 --- **
  4. Simplify constants** -16 + 13 = -3 So: x² + 8x + 13 = (x + 4)² - 3 --- **Final Answer:** (x + 4)² - 3

Let's solve the problem step-by-step. We start with: x² + 8x + 13 --- **

5 x² - 10x + 24 = 0

Hint: To solve quadratic equations by completing the square, move the constant term to the other side, then add the square of half the coefficient of x to both sides to create a perfect square trinomial.

Show the answer

Answer: x = 4, 6

  1. Start with x² - 10x + 24 = 0
  2. Move the constant term to the right side: x² - 10x = -24
  3. Take half of the coefficient of x: -10 ÷ 2 = -5
  4. Square this result: (-5)² = 25
  5. Add 25 to both sides: x² - 10x + 25 = -24 + 25
  6. Simplify: x² - 10x + 25 = 1
  7. Factor the left side as a perfect square: (x - 5)² = 1
  8. Take square root of both sides: x - 5 = ±1
  9. Solve for x: x = 5 + 1 = 6 or x = 5 - 1 = 4 The solutions are x = 4 and x = 6.

6 x² - 6x + 5 = (x - ?)² - 4

Hint: Focus on the coefficient of the x-term and consider how completing the square transforms the expression

Show the answer

Answer: 3

  1. Start with the original expression: x² - 6x + 5
  2. To complete the square, take half of the coefficient of x: -6/2 = -3
  3. Square this value: (-3)² = 9
  4. Add and subtract 9: x² - 6x + 9 - 9 + 5
  5. Group the perfect square trinomial: (x² - 6x + 9) + (-9 + 5)
  6. Write as a squared binomial: (x - 3)² + (-4)
  7. Compare with the given form: (x - 3)² - 4
  8. The missing value is 3

The answer is 3.

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