Quadratic Formula: Derive Standard Form

Grade 9 · algebra · 71 practice problems · read aloud

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The Quadratic Formula

The quadratic formula is a powerful tool that solves any quadratic equation. A quadratic equation is an equation where the highest power of the variable (usually 'x') is 2. They look like: ax² + bx + c = 0.

It's useful because methods like factoring don't always work. The quadratic formula always gives you the solution(s)!

The Formula

The solutions for x in ax² + bx + c = 0 are:

x = [-b ± √(b² - 4ac)] / (2a)

Step-by-Step Guide

  1. Identify a, b, and c. Write down the numbers in front of x², x, and the constant.
  2. Plug into the formula. Carefully substitute a, b, and c into the formula.
  3. Simplify under the radical. Calculate the value of b² - 4ac (this is the discriminant).
  4. Solve for both answers. Calculate the + version and the - version.

Example 1: Two Real Solutions

Solve: x² + 6x + 8 = 0

  1. a = 1, b = 6, c = 8
  2. x = [-6 ± √(6² - 4(1)(8))] / (2(1))
  3. x = [-6 ± √(36 - 32)] / 2
  4. x = [-6 ± √4] / 2
  5. x = (-6 + 2)/2 = -2 OR x = (-6 - 2)/2 = -4

Solutions: x = -2, -4

Example 2: One Real Solution

Solve: 2x² - 4x + 2 = 0

  1. a = 2, b = -4, c = 2
  2. x = [4 ± √((-4)² - 4(2)(2))] / (2(2))
  3. x = [4 ± √(16 - 16)] / 4
  4. x = [4 ± √0] / 4
  5. x = 4 / 4 = 1

Solution: x = 1

Common Mistakes ⚠️

  • Sign Errors: 'b' is the coefficient, including its sign. If the equation is x² - 3x + 2 = 0, then b = -3.
  • Order of Operations: Simplify under the square root (b² - 4ac) completely before taking the square root.
  • The Denominator: The entire numerator, -b ± √(b²-4ac), is divided by 2a. Use parentheses!
  • Negative Coefficients: When b is negative, -b becomes positive. For example, if b = -5, then -b = -(-5) = 5.

Tips & Tricks

  • Memory Aid: Sing to "Pop Goes the Weasel": "x equals negative b, plus or minus the square root, of b squared minus four a c, all over two a!" 🎵
  • Discriminant Check: Look at b² - 4ac first.
    • Positive = Two real solutions
    • Zero = One real solution
    • Negative = Two complex solutions (for later grades!)
  • Always write the equation in standard form (ax²+bx+c=0) first.

How to Practice

Start with simple equations where a=1. Then move to equations with a > 1 and negative values for b and c. Check your answers by plugging them back into the original equation. Practice at least 3-5 problems a day to build muscle memory. The more you use it, the easier it gets!

Practice problems

6 of the 71, worked through step by step — try them before opening the answer.

1 x² + 5x + 6 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term.

Show the answer

Answer: x = -2, -3

  1. Identify the coefficients. The equation is in the form ax² + bx + c = 0. Here, a = 1, b = 5, c = 6.
  2. Factor the quadratic expression. We look for two numbers that multiply to a * c = 1 * 6 = 6, and add to b = 5. The numbers 2 and 3 multiply to 6 and add to 5.
  3. Rewrite the middle term using these numbers. x² + 5x + 6 = x² + 2x + 3x + 6.
  4. Factor by grouping. Group the terms: (x² + 2x) + (3x + 6). Factor each group: x(x + 2) + 3(x + 2).
  5. Factor out the common binomial (x + 2). (x + 2)(x + 3) = 0.
  6. Apply the zero-product property. If (x + 2)(x + 3) = 0, then: x + 2 = 0 or x + 3 = 0.
  7. Solve each equation. x + 2 = 0 → x = -2 x + 3 = 0 → x = -3 Final answer: x = -2, -3

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

2 x² + 4x - 5 = 0

Hint: This is a quadratic equation that can be solved using the quadratic formula. Remember to identify the coefficients a, b, and c first.

Show the answer

Answer: x = 1, -5

  1. Identify coefficients: a = 1, b = 4, c = -5
  2. Apply quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a)
  3. Calculate discriminant: b² - 4ac = 4² - 4(1)(-5) = 16 + 20 = 36
  4. Calculate square root: sqrt(36) = 6
  5. Apply formula: x = [-4 ± 6] / (2×1)
  6. First solution: x = (-4 + 6)/2 = 2/2 = 1
  7. Second solution: x = (-4 - 6)/2 = -10/2 = -5 The solutions are x = 1 and x = -5.

3 Derive x = [-b ± √(b²-4ac)] / 2a by completing square on ax²+bx+c=0

Hint: Start by isolating the quadratic and linear terms, then find the value needed to complete the square for the general form

Show the answer

Answer: x = [-b ± √(b²-4ac)] / 2a

  1. Start with the general quadratic equation: ax² + bx + c = 0
  2. Divide through by a: x² + (b/a)x + c/a = 0
  3. Move constant term to right side: x² + (b/a)x = -c/a
  4. Complete the square: Take half of (b/a), square it to get (b/2a)² = b²/4a²
  5. Add this to both sides: x² + (b/a)x + b²/4a² = -c/a + b²/4a²
  6. Write left side as perfect square: (x + b/2a)² = -c/a + b²/4a²
  7. Combine right side over common denominator: (x + b/2a)² = (-4ac + b²)/4a²
  8. Take square root of both sides: x + b/2a = ±√(b² - 4ac)/2a
  9. Solve for x: x = [-b ± √(b² - 4ac)] / 2a

4 Derive x = [-b ± √(b²-4ac)] / (2a) by completing the square on ax² + bx + c = 0

Hint: Start by isolating the quadratic and linear terms, then find the value needed to complete the square for the general form.

Show the answer

Answer: x = [-b ± √(b²-4ac)] / (2a)

  1. Start with the general quadratic equation: ax² + bx + c = 0
  2. Subtract c from both sides: ax² + bx = -c
  3. Divide through by a: x² + (b/a)x = -c/a
  4. Complete the square by adding (b/(2a))² to both sides: x² + (b/a)x + (b/(2a))² = -c/a + (b/(2a))²
  5. Write left side as perfect square: (x + b/(2a))² = -c/a + b²/(4a²)
  6. Combine right side terms: (x + b/(2a))² = (-4ac + b²)/(4a²)
  7. Take square root of both sides: x + b/(2a) = ±√(b² - 4ac)/(2a)
  8. Isolate x: x = [-b ± √(b² - 4ac)]/(2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).

5 Derive x = [-b ± √(b²-4ac)] / (2a) by completing the square on ax² + bx + c = 0.

Hint: Start by moving the constant term to the right side, then divide through by the coefficient of x². Think about what number you need to add to both sides to make the left side a perfect square trinomial.

Show the answer

Answer: x = [-b ± √(b²-4ac)] / (2a)

  1. Start with the general quadratic equation: ax² + bx + c = 0
  2. Subtract c from both sides: ax² + bx = -c
  3. Divide both sides by a: x² + (b/a)x = -c/a
  4. Complete the square. Take half of the coefficient of x: (b/a) ÷ 2 = b/(2a). Square it: (b/(2a))² = b²/(4a²). Add this to both sides: x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²)
  5. Write the left side as a perfect square: (x + b/(2a))² = -c/a + b²/(4a²)
  6. Combine the right side over a common denominator: (x + b/(2a))² = (-4ac + b²)/(4a²)
  7. Take the square root of both sides: x + b/(2a) = ±√(b² - 4ac)/(2a)
  8. Isolate x by subtracting b/(2a) from both sides: x = -b/(2a) ± √(b² - 4ac)/(2a)
  9. Combine the fractions: x = [-b ± √(b² - 4ac)] / (2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).

6 Derive the quadratic formula x = [-b ± √(b²-4ac)] / (2a) by completing the square on ax² + bx + c = 0.

Hint: Think about how to transform the left side into a perfect square trinomial. Start by moving the constant term to the right, then divide by the coefficient of x². What must you add to both sides to create a perfect square?

Show the answer

Answer: x = [-b ± √(b²-4ac)] / (2a)

  1. Start with the general quadratic equation: ax² + bx + c = 0
  2. Subtract c from both sides: ax² + bx = -c
  3. Divide both sides by a (a ≠ 0): x² + (b/a)x = -c/a
  4. Complete the square. Take half of the coefficient of x: (b/a) ÷ 2 = b/(2a). Square it: (b/(2a))² = b²/(4a²). Add this to both sides: x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²)
  5. Write the left side as a perfect square: (x + b/(2a))² = -c/a + b²/(4a²)
  6. Combine the right side over a common denominator 4a²: (x + b/(2a))² = (-4ac + b²)/(4a²) = (b² - 4ac)/(4a²)
  7. Take the square root of both sides: x + b/(2a) = ±√(b² - 4ac)/(2a)
  8. Isolate x by subtracting b/(2a) from both sides: x = -b/(2a) ± √(b² - 4ac)/(2a)
  9. Combine the terms over the common denominator 2a: x = [-b ± √(b² - 4ac)]/(2a) The quadratic formula is derived as x = [-b ± √(b²-4ac)]/(2a).
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