Quadratic Formula

Core Definition

The Quadratic Formula is a universal algebraic formula used to solve any second-degree polynomial equation of the form $ax^2 + bx + c = 0$ (where $a \neq 0$ ). The solutions, or roots, are given by:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Concept Explanation

The Quadratic Formula is derived directly by completing the square on the general equation $ax^2 + bx + c = 0$ . It guarantees that you can find the roots of any quadratic equation, even when the expression is not factorable using rational integers.

The Discriminant

The expression under the radical, $b^2 - 4ac$ , is called the discriminant (often denoted by the Greek letter delta, $\Delta$ or $D$ ). The value of the discriminant determines the nature and number of the roots:

Discriminant Value	Graphical Interpretation	Nature of the Roots
$b^2 - 4ac > 0$	Parabola crosses the $x$ -axis twice.	Two distinct real roots.
$b^2 - 4ac = 0$	Parabola touches the $x$ -axis at its vertex.	One repeated real root.
$b^2 - 4ac < 0$	Parabola lies entirely above or below the $x$ -axis.	Two complex conjugate roots ( $x = \alpha \pm \beta i$ ).

When to Use the Quadratic Formula vs. Factoring

Step 1: Check for easy factoring. If the coefficients are small and $a=1$ , try to find factors of $c$ that add to $b$ . If you can spot them in a few seconds, factoring is faster.
Step 2: Check the discriminant.
- If $b^2 - 4ac$ is a perfect square (e.g. $1, 4, 9, 16, 25 \dots$ ), the quadratic is factorable over rational numbers.
- If $b^2 - 4ac$ is not a perfect square (e.g. $5, 12, 17$ ), the quadratic cannot be factored over integers. You must use the Quadratic Formula or complete the square. The roots will be irrational numbers containing radicals.
- If $b^2 - 4ac$ is negative, the roots are complex numbers.

How to Recognize When to Use It in Calculus

In calculus, you will encounter quadratics in several key contexts where the Quadratic Formula is indispensable:

1. Identifying Irreducible Quadratics (Integration)

When integrating rational functions using Partial Fraction Decomposition, you must factor the denominator. If a quadratic factor in the denominator has a negative discriminant ( $b^2 - 4ac < 0$ ), it is irreducible. You cannot factor it further, and you must integrate it using completing the square and the inverse tangent formula:

\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C

2. Finding Critical Points (Curve Sketching)

To find local extrema (peaks and valleys) of a cubic function $f(x) = Ax^3 + Bx^2 + Cx + D$ , you take the first derivative, which is a quadratic: $f'(x) = 3Ax^2 + 2Bx + C$ . Setting $f'(x) = 0$ and solving for $x$ via the Quadratic Formula yields the critical values.

Example Problems

Example Problem 1: Two Irrational Roots

Solve the equation:

x^2 - 4x + 1 = 0

Solution

Step 1: Identify coefficients $a = 1, b = -4, c = 1$ .

Step 2: Calculate the discriminant

D = b^2 - 4ac = (-4)^2 - 4(1)(1) = 16 - 4 = 12

Since

D = 12 > 0

, there are two real roots. Because

12

is not a perfect square, these roots are irrational and cannot be found by simple factoring.

Step 3: Apply the Quadratic Formula

x = \frac{-(-4) \pm \sqrt{12}}{2(1)} = \frac{4 \pm \sqrt{4 \cdot 3}}{2} = \frac{4 \pm 2\sqrt{3}}{2}

Step 4: Simplify Factor out a $2$ from the numerator to cancel the denominator:

x = \frac{2(2 \pm \sqrt{3})}{2} = 2 \pm \sqrt{3}

The solutions are $x = 2 + \sqrt{3}$ and $x = 2 - \sqrt{3}$ .

Example Problem 2: Irreducible / Complex Roots

Solve the equation:

x^2 + 2x + 5 = 0

Solution

Step 1: Identify coefficients $a = 1, b = 2, c = 5$ .

Step 2: Calculate the discriminant

D = b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16

Since

D = -16 < 0

, the roots are complex conjugate pairs, and the expression

x^2 + 2x + 5

is an irreducible quadratic.