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The Quadratic Formula and Discriminant

A free Algebra I lesson from the “Quadratic Foundations” unit, with a worked example and practice problems including step-by-step solutions.

For ax^2 + bx + c = 0 with a != 0, completing the square leads to (2ax + b)^2 = b^2 - 4ac. Taking both square-root branches and dividing by 2a gives x = (-b + sqrt(b^2 - 4ac))/(2a) or x = (-b - sqrt(b^2 - 4ac))/(2a). This is the quadratic formula, so its two signs come from the same square-root reasoning used when completing the square. Write the equation with zero on one side before identifying a, b, and c, including the signs of the coefficients. The discriminant D = b^2 - 4ac determines the number of distinct real roots: two if D > 0, one if D = 0, and none if D < 0. With integer coefficients, a positive discriminant that is not a perfect square produces irrational roots. Keep radicals exact during calculation; round only when a requested measurement or approximation requires it. The whole numerator is divided by 2a. A real algebraic root may still be outside a model's physical domain, such as a negative length.

What you'll learn

Why it matters: Area and motion equations often have irrational solutions. Exact roots preserve mathematical information; approximations help express practical measurements.

Worked example

Problem. Solve 2x^2 - 4x - 1 = 0. Give exact roots and approximations to three decimal places.

  1. The coefficients are a = 2, b = -4, and c = -1.
  2. D = (-4)^2 - 4(2)(-1) = 16 + 8 = 24; because D is positive, there are two distinct real roots.
  3. Substitute: x = (4 + sqrt(24))/4 or (4 - sqrt(24))/4.
  4. Since sqrt(24) = 2sqrt(6), divide every numerator term by 4 to obtain x = 1 + sqrt(6)/2 or 1 - sqrt(6)/2.
  5. Using sqrt(6) approximately 2.449490 gives roots approximately 2.225 and -0.225. The exact expressions remain the solutions before rounding.
  6. Completing the square gives (x - 1)^2 = 3/2; both exact roots satisfy that equivalent equation.

Answer: x = 1 + sqrt(6)/2 or x = 1 - sqrt(6)/2; approximately 2.225 or -0.225.

Lesson preview: this page shows 5 practice items. The interactive activity contains 32 practice items and 17 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Which pair gives the exact roots of x^2 - 2x - 2 = 0?

Choices: 1 + sqrt(3) and 1 - sqrt(3) · 2 + sqrt(3) and 2 - sqrt(3) · 1 + sqrt(2) and 1 - sqrt(2) · 3 and -1

Show solution
  1. D = (-2)^2 - 4(1)(-2) = 12.
  2. x = (2 + sqrt(12))/2 or (2 - sqrt(12))/2.
  3. Use sqrt(12) = 2sqrt(3) to obtain 1 + sqrt(3) and 1 - sqrt(3).

Answer: 1 + sqrt(3) and 1 - sqrt(3)

2. For 3x^2 + 2x = 5, compute the discriminant after writing standard form.

Show solution
  1. Standard form is 3x^2 + 2x - 5 = 0, so c = -5.
  2. D = 2^2 - 4(3)(-5) = 4 + 60 = 64.

Answer: 64

3. A student writes (6 + sqrt(20))/2 = 3 + sqrt(20). What is the correction?

Choices: Divide the radical term by 2 as well, giving 3 + sqrt(5). · Divide only the 6 because radicals cannot be divided. · Replace sqrt(20) with 10. · Multiply the numerator by 2.

Show solution
  1. Both numerator terms are divided by 2: 6/2 + sqrt(20)/2.
  2. Since sqrt(20) = 2sqrt(5), the result is 3 + sqrt(5).

Answer: Divide the radical term by 2 as well, giving 3 + sqrt(5).

4. A rectangle is 2 meters longer than it is wide and has area 10 square meters. Which is its width?

Choices: -1 + sqrt(11) meters · -1 - sqrt(11) meters · 1 + sqrt(11) meters · 5 meters

Show solution
  1. Let w be the positive width. Then w(w + 2) = 10, or w^2 + 2w - 10 = 0.
  2. The quadratic formula gives w = (-2 + sqrt(44))/2 or (-2 - sqrt(44))/2, equal to -1 + sqrt(11) or -1 - sqrt(11).
  3. Only -1 + sqrt(11) is positive, so it is the physical width.

Answer: -1 + sqrt(11) meters

5. Solve x^2 - 5x + 6 = 0 by the quadratic formula. Enter the larger solution.

Show solution
  1. a=1, b=-5, c=6; discriminant 25 - 24 = 1.
  2. x = (5 plus or minus 1)/2 = 3 or 2.
  3. The larger is 3.

Answer: 3

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Derive, check, and interpret exact roots

Solve x^2 - 6x + 4 = 0 using the quadratic formula. Keep both roots exact, verify them by completing the square, and give each to two decimal places. Explain why rounding before the last step is unnecessary.

Review rubric and solution
  • Calculate D = 20 and obtain roots 3 + sqrt(5) and 3 - sqrt(5).
  • Use (x - 3)^2 = 5 to verify both roots.
  • Give approximations 5.24 and 0.76.
  • Explain that exact radical values avoid accumulated rounding changes.
  1. The formula gives x = (6 plus or minus sqrt(20))/2 = 3 plus or minus sqrt(5).
  2. The original equation is equivalent to x^2 - 6x + 9 = 5, or (x - 3)^2 = 5, which both exact roots satisfy.
  3. Since sqrt(5) is approximately 2.236068, the roots round to 5.24 and 0.76.
  4. Keeping the radical exact until the final approximation avoids changing intermediate quantities by rounding.

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