Unit 9 · Lesson 3 of 5

The quadratic formula and the discriminant

About 25 minutes · 5 steps · one idea: one formula solves every quadratic

You'll be able to: solve any quadratic equation with the quadratic formula, simplify the answer, and tell from the discriminant how many real solutions there are.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. The second one is the heart of today's lesson.

Solve by completing the square: \(x^2 + 6x + 4 = 0\).

Move the constant: \(x^2 + 6x = -4\). Half of 6 is 3, and \(3^2 = 9\). Add 9: \(x^2 + 6x + 9 = 5\).

So \((x + 3)^2 = 5\), \(x + 3 = \pm\sqrt{5}\), and \(x = -3 \pm \sqrt{5}\).

Evaluate \(b^2 - 4ac\) when \(a = 2\), \(b = -3\), and \(c = -5\).

Use parentheses: \((-3)^2 - 4(2)(-5) = 9 - (-40) = 9 + 40\).

Answer: \(49\).

Step 2 of 5The idea

Completing the square always works, but it takes many steps. If you complete the square on \(ax^2 + bx + c = 0\) using letters instead of numbers, you get one formula that does all of those steps at once. It is the quadratic formula.

First the equation must be in standard form: everything on one side, 0 on the other. Then read off \(a\), \(b\), and \(c\), each with its sign.

Quadratic formula. The solutions of \(ax^2 + bx + c = 0\) are \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

Discriminant. The part under the root, \(b^2 - 4ac\), tells you what kind of solutions to expect:

positive: two real solutions · zero: one real solution · negative: two complex solutions (no real ones).

Simplify. Simplify the radical first. Then, if every term on top shares a factor with the bottom, divide all of them by it.

The discriminant is worth computing first. It catches mistakes early and tells you whether you'll need \(i\).

Why this shows up on the placement test: many quadratics on the test do not factor, and "how many real solutions?" is a classic quick question.

Step 3 of 5Worked examples

Example 1: \(x^2 = 4x + 1\)

  1. Standard form: subtract \(4x\) and \(1\) from both sides: \(x^2 - 4x - 1 = 0\). So \(a = 1\), \(b = -4\), \(c = -1\).
  2. Discriminant: \((-4)^2 - 4(1)(-1) = 16 + 4 = 20\). Positive, so two real solutions.
  3. Formula: \(x = \dfrac{-(-4) \pm \sqrt{20}}{2(1)} = \dfrac{4 \pm \sqrt{20}}{2}\).
  4. Simplify the radical: \(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\), so \(x = \dfrac{4 \pm 2\sqrt{5}}{2}\).
  5. Divide each term on top by 2: \(x = 2 \pm \sqrt{5}\).

Example 2: \(3x^2 - 2x + 1 = 0\)

  1. Read the numbers: \(a = 3\), \(b = -2\), \(c = 1\).
  2. Discriminant: \((-2)^2 - 4(3)(1) = 4 - 12 = -8\). Negative, so two complex solutions.
  3. Formula: \(x = \dfrac{2 \pm \sqrt{-8}}{6}\).
  4. Simplify the radical: \(\sqrt{-8} = i\sqrt{8} = 2i\sqrt{2}\), so \(x = \dfrac{2 \pm 2i\sqrt{2}}{6}\).
  5. Divide every term by 2: \(x = \dfrac{1 \pm i\sqrt{2}}{3}\). You can only cancel the 2 because both terms on top have it.

Step 4 of 5Practice

Type your answer and press Check. For equations, list every solution separated by commas, or use ± (type +- or ±). If you miss twice, open the worked solution. Then mark how it went.

Find the discriminant of \(2x^2 - 5x + 4 = 0\).

\(a = 2\), \(b = -5\), \(c = 4\).

\(b^2 - 4ac = (-5)^2 - 4(2)(4) = 25 - 32 = -7\).

Answer: \(-7\) (negative, so two complex solutions).

Which equation has two complex (non-real) solutions?

  1. \(x^2 + 4x - 5 = 0\)
  2. \(x^2 + 4x + 5 = 0\)
  3. \(x^2 + 4x + 4 = 0\)
  4. \(x^2 + 5x + 4 = 0\)

Find each discriminant \(b^2 - 4ac\). Complex solutions need a negative one.

(a) \(16 - 4(1)(-5) = 36\). (b) \(16 - 4(1)(5) = -4\). (c) \(16 - 16 = 0\). (d) \(25 - 16 = 9\).

Only (b) is negative. Answer: (b).

Solve with the quadratic formula: \(x^2 + 5x + 2 = 0\).

\(a = 1\), \(b = 5\), \(c = 2\). Discriminant: \(25 - 8 = 17\).

\(x = \dfrac{-5 \pm \sqrt{17}}{2}\). Nothing simplifies further.

Solve with the quadratic formula: \(2x^2 - 7x + 3 = 0\).

\(a = 2\), \(b = -7\), \(c = 3\). Discriminant: \(49 - 24 = 25\), and \(\sqrt{25} = 5\).

\(x = \dfrac{7 \pm 5}{4}\): \(x = \dfrac{12}{4} = 3\) or \(x = \dfrac{2}{4} = \dfrac{1}{2}\).

Solve: \(2x^2 = 4x + 3\).

Standard form: \(2x^2 - 4x - 3 = 0\), so \(a = 2\), \(b = -4\), \(c = -3\).

Discriminant: \(16 - 4(2)(-3) = 16 + 24 = 40\), and \(\sqrt{40} = 2\sqrt{10}\).

\(x = \dfrac{4 \pm 2\sqrt{10}}{4}\). Divide every term by 2: \(x = \dfrac{2 \pm \sqrt{10}}{2}\).

Solve: \(x^2 - 4x + 7 = 0\).

\(a = 1\), \(b = -4\), \(c = 7\). Discriminant: \(16 - 28 = -12\).

\(\sqrt{-12} = i\sqrt{12} = 2i\sqrt{3}\), so \(x = \dfrac{4 \pm 2i\sqrt{3}}{2}\).

Divide each term by 2: \(x = 2 \pm i\sqrt{3}\).

Solve: \(x(2x + 1) = 4\).

Multiply out and get 0 on one side: \(2x^2 + x - 4 = 0\). So \(a = 2\), \(b = 1\), \(c = -4\).

Discriminant: \(1 - 4(2)(-4) = 1 + 32 = 33\).

\(x = \dfrac{-1 \pm \sqrt{33}}{4}\).

A rectangle is 3 feet longer than it is wide, and its area is 20 square feet. Find its width exactly.

Let the width be \(w\). Then the length is \(w + 3\), and \(w(w + 3) = 20\), so \(w^2 + 3w - 20 = 0\).

Discriminant: \(9 - 4(1)(-20) = 9 + 80 = 89\). So \(w = \dfrac{-3 \pm \sqrt{89}}{2}\).

\(\sqrt{89}\) is a bit more than 9, so the "minus" answer is negative. A width can't be negative.

Answer: \(w = \dfrac{-3 + \sqrt{89}}{2}\) feet (about 3.2 feet).

Step 5 of 5Wrap-up

You can now solve any quadratic: put it in standard form, read \(a\), \(b\), \(c\), check the discriminant, use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and simplify.

Read or watch next: OpenStax Intermediate Algebra 2e, 9.3 Solve Quadratic Equations Using the Quadratic Formula (free textbook, with more practice at the end of the section).

Also good: Khan Academy: The quadratic formula (short videos). Quick rules: Quadratics cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "How do I know when to factor and when to use the quadratic formula?" It knows this course and can explain things another way.