Quadratic formula

High School

Definition

The most common formula to solve quadratic equations. The formula can be used to quicly determine how many real zeros the equation has.

Worked examples

\(x^2 + 5x + 6 = 0 \)→\( x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2}\)

Identify \(a=1\), \(b=5\), \(c=6\), substitute into \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), then simplify to get \(x = -2\) or \(x = -3\).

\(2x^2 - 3x - 5 = 0 \)→\( x = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4}\)

With \(a=2\), \(b=-3\), \(c=-5\), the discriminant \(49\) is positive, so two real solutions: \(x = 2.5\) or \(x = -1\).

\(x^2 + 4x + 5 = 0 \)→\( x = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}\)

The discriminant \(-4\) is negative, so there are no real zeros — the parabola does not cross the x-axis.

Common mistakes

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2}\) → \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Forgetting the 'a' in the denominator gives wrong solutions; divide by \(2a\), not just \(2\).
For \(3x^2 - 2x + 1 = 0\), using \(b = 2\) instead of \(b = -2\) → \(b = -2\) (keep the sign from the equation) The sign of \(b\) and \(c\) must match the standard form \(ax^2 + bx + c = 0\).
\(\sqrt{b^2 - 4ac} = b - 2\sqrt{ac}\) → \(\sqrt{b^2 - 4ac}\) stays as one square root The discriminant is a single value under the radical; you cannot split it into separate roots.

Where you'll use it next

You'll use the quadratic formula throughout algebra to solve any quadratic equation, analyze parabolas, and find roots in calculus, physics (projectile motion), and engineering applications.

Found in 1 StudyPug lesson

The Quadratic Formula

Grade 11 Math

The quadratic formula solves any equation of the form a x squared plus b x plus c equals zero.

11th Grade11th