Quadratic Formula

Learn how to solve any quadratic equation using the quadratic formula, with clear steps, examples, discriminant use, and common mistakes students should avoid.

1. Introduction

The quadratic formula is a universal method to solve any quadratic equation of the form:

\(ax^2 + bx + c = 0\)

It works even when factorisation or completing the square is difficult. The formula gives the solutions (roots) directly by substituting the values of a, b, and c.

2. The Quadratic Formula

The solutions of the quadratic equation:

\(ax^2 + bx + c = 0\)

are given by:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

The symbol \(\pm\) means that there are generally two solutions:

  • \(x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}\)
  • \(x = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}\)

3. Meaning of Each Part of the Formula

The quadratic formula contains three main components:

  • \(-b\): Opposite of the linear coefficient
  • \(b^2 - 4ac\): Discriminant (decides nature of roots)
  • \(2a\): Denominator that normalises the expression

The term \(b^2 - 4ac\) appears under the square root and plays a key role in determining whether the roots are real or imaginary.

4. Steps to Apply the Quadratic Formula

To solve an equation using the quadratic formula:

  1. Write the equation in standard form: \(ax^2 + bx + c = 0\).
  2. Identify the values of a, b, and c.
  3. Substitute them into the formula:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

  1. Simplify inside the square root.
  2. Evaluate both the + and - cases to find the two roots.

4.1. Tip for Students

Always rewrite the equation in standard form first. If the equation is not arranged properly, a, b, and c may be identified incorrectly.

5. Worked Examples

These examples show how to use the quadratic formula step-by-step:

5.1. Example 1: Simple Numbers

Solve: \(x^2 - 5x + 6 = 0\)

Here: \(a = 1\), \(b = -5\), \(c = 6\)

Discriminant: \(b^2 - 4ac = 25 - 24 = 1\)

\(x = \dfrac{-(-5) \pm \sqrt{1}}{2} = \dfrac{5 \pm 1}{2}\)

Solutions: \(x = 3\), \(x = 2\)

5.2. Example 2: Roots Involving Irrational Numbers

Solve: \(x^2 - 3x - 2 = 0\)

\(a = 1, b = -3, c = -2\)

Discriminant: \(9 + 8 = 17\)

\(x = \dfrac{3 \pm \sqrt{17}}{2}\)

5.3. Example 3: When a ≠ 1

Solve: \(2x^2 + 3x - 5 = 0\)

\(a = 2, b = 3, c = -5\)

Discriminant: \(9 + 40 = 49\)

\(x = \dfrac{-3 \pm 7}{4}\)

Solutions: \(x = 1\), \(x = -\dfrac{5}{2}\)

5.4. Example 4: When Roots Are Equal

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

\(a = 1, b = -4, c = 4\)

Discriminant: \(16 - 16 = 0\)

So both roots are:

\(x = \dfrac{4}{2} = 2\)

6. Connection to the Discriminant

The discriminant \(D = b^2 - 4ac\) tells us about the nature of roots:

  • If \(D > 0\): two real and distinct roots
  • If \(D = 0\): one real repeated root
  • If \(D < 0\): no real roots (complex roots)

7. Advantages of Using the Quadratic Formula

  • Works for all quadratic equations.
  • No need for factorisation.
  • Efficient when numbers are large or roots are complex.
  • Direct substitution gives exact solutions.

8. Common Mistakes

  • Incorrect identification of a, b, and c.
  • Forgetting to include ± in the formula.
  • Incorrect simplification of the discriminant.
  • Sign errors when computing −b.
  • Dividing incorrectly by 2a.

9. Quick Practice

Solve the following using the quadratic formula:

  1. \(x^2 + 2x - 3 = 0\)
  2. \(3x^2 - x - 4 = 0\)
  3. \(x^2 + x + 1 = 0\)
  4. \(5x^2 - 6x + 1 = 0\)

10. Summary

  • The quadratic formula gives the solutions of any quadratic equation.
  • Roots are computed using: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Discriminant decides the nature of the roots.
  • Always simplify carefully to avoid sign errors.