Quadratic Equations

A quadratic equation is any equation you can write in the form

$ax^2 + bx + c = 0, \quad a \neq 0$

The defining feature is the $x^2$ term — the variable appears squared, and that single detail changes everything. A linear equation has one solution because a straight line crosses zero once. A quadratic can have two solutions, one, or none (among real numbers), because its graph is a parabola — a U-shaped curve that can cross the horizontal axis twice, touch it once, or miss it entirely.

Learning Objectives

By the end of this page, you should be able to:

• Recognize a quadratic equation and identify its coefficients $a$, $b$, and $c$
• Solve quadratics by three methods — factoring, completing the square, and the quadratic formula — and choose the best method for a given problem
• Use the discriminant to predict how many real solutions exist before solving
• Connect the algebra to the geometry of the parabola (roots, vertex, axis of symmetry)
• Set up and solve real-world problems that reduce to a quadratic

Quick Answer

To solve $ax^2 + bx + c = 0$, you can always use the quadratic formula:

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

It never fails. Faster methods exist when the numbers cooperate — factoring if the expression splits neatly, and completing the square when you need the vertex — but the formula is the universal tool. The quantity under the root, $b^2 - 4ac$, is the discriminant: if it's positive there are two real solutions, if zero there's exactly one, and if negative there are none among the real numbers.

Why Quadratics Exist

Quadratics appear the moment two quantities multiply to produce a third — which happens constantly in the real world. The area of a rectangle is length times width; if both depend on the same variable, you get a quadratic. A ball thrown into the air follows a parabolic path because gravity acts on distance-per-time-squared. Profit is often revenue minus cost, and when price affects both quantity sold and revenue, the result is quadratic. Understanding quadratics means understanding the mathematics of area, projectile motion, and optimization.

Where It Came From

Quadratics are one of the oldest problems in mathematics, and they were born from an intensely practical need: land and area. Around 2000 BC, Babylonian scribes were dividing fields, calculating grain storage, and settling inheritance disputes — all problems of the form "I know the area and I know a relationship between the sides; find the sides." That is a quadratic. Clay tablets survive showing the Babylonians solving these by a recipe equivalent to completing the square, though they had no symbols and worked entirely in words and specific numbers.

The Greeks, distrustful of the not-yet-understood irrational numbers, recast quadratics geometrically — literally as areas of squares and rectangles, which is why we still say "completing the square" and "$x$ squared." The step toward a general algebraic solution came from India: Brahmagupta (628 AD) gave an explicit rule close to the modern formula and, crucially, accepted negative and zero quantities that the Greeks had rejected.

The word "algebra" itself comes from the 9th-century Persian mathematician al-Khwārizmī, whose book al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wa'l-muqābala gave systematic procedures for solving quadratics; al-jabr ("restoring", i.e. moving a term to the other side) became "algebra." The compact formula we use today, with its $\pm$ and square-root sign, only became possible after the symbolic notation developed in 16th–17th century Europe. In short: the problem is 4,000 years old, but the tidy formula is only about 400 years old — a reminder that good notation is itself a hard-won invention.

The Three Methods

1. Factoring

If you can rewrite $ax^2 + bx + c$ as a product of two linear factors, the zero-product property finishes the job: if two things multiply to zero, at least one of them is zero.

Example: Solve $x^2 - 5x + 6 = 0$.

Find two numbers that multiply to $+6$ (the constant) and add to $-5$ (the middle coefficient): those are $-2$ and $-3$.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

So $x - 2 = 0$ or $x - 3 = 0$, giving $x = 2$ or $x = 3$.

Factoring is fastest when it works, but many quadratics don't factor with tidy integers. Don't waste minutes hunting for factors that aren't there — switch methods.

2. The Quadratic Formula

This is the method that always works. For $ax^2 + bx + c = 0$:

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

Example: Solve $2x^2 + 3x - 5 = 0$. Here $a = 2$, $b = 3$, $c = -5$.

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}$

So $x = \frac{4}{4} = 1$ or $x = \frac{-10}{4} = -2.5$.

The $\pm$ symbol is why quadratics give two answers: you compute once with $+$ and once with $-$.

3. Completing the Square

This method rewrites the equation so the variable appears only once, inside a perfect square. It's the technique that derives the quadratic formula, and it's the fastest way to find a parabola's vertex.

Example: Solve $x^2 + 6x + 5 = 0$.

Take the coefficient of $x$ (which is 6), halve it to get 3, and square it to get 9. Add and subtract 9:

$x^2 + 6x + 9 - 9 + 5 = 0 \implies (x + 3)^2 - 4 = 0 \implies (x+3)^2 = 4$

Take the square root of both sides: $x + 3 = \pm 2$, so $x = -1$ or $x = -5$.

The Discriminant: Predicting Solutions

Before solving, the discriminant $D = b^2 - 4ac$ tells you what to expect:

Discriminant	Real solutions	Parabola and x-axis
$D > 0$	Two distinct real solutions	Crosses the x-axis at two points
$D = 0$	One real solution (repeated)	Just touches the x-axis at the vertex
$D < 0$	No real solutions	Never reaches the x-axis

This is a genuine time-saver on exams: if $D < 0$, stop — there are no real answers to find.

The Geometry: The Parabola

Every quadratic $y = ax^2 + bx + c$ graphs as a parabola.

• If $a > 0$, it opens upward (a valley); if $a < 0$, it opens downward (a hill).
• The solutions (roots) are where the parabola crosses the x-axis — exactly the values that make $y = 0$.
• The vertex (the turning point) sits at $x = -\dfrac{b}{2a}$, on the axis of symmetry. Completing the square reveals it directly.

Seeing the graph and the algebra together is the "aha" of quadratics: two roots means two crossings, one root means a single touch, no real roots means the curve floats entirely above or below the axis.

Real-World Application

Projectile problem: A ball is thrown upward; its height in metres after $t$ seconds is $h = -5t^2 + 20t$. When does it hit the ground ($h = 0$)?

$-5t^2 + 20t = 0 \implies -5t(t - 4) = 0 \implies t = 0 \text{ or } t = 4$

It leaves the ground at $t = 0$ and lands at $t = 4$ seconds. The vertex at $t = -\frac{20}{2(-5)} = 2$ tells you it reaches maximum height at 2 seconds — exactly halfway, by symmetry.

Common Mistakes

Mistake 1: Forgetting the $\pm$. Students often compute only the $+$ branch of the quadratic formula and report one answer. A quadratic normally has two solutions — always evaluate both signs.

Mistake 2: Sign errors with $b$ and $c$. In $2x^2 - 3x - 5 = 0$, it's easy to plug in $b = 3$ instead of $b = -3$. Write the coefficients out explicitly, including their signs, before substituting.

Mistake 3: Dividing by $x$ to "simplify". Given $x^2 = 5x$, dividing both sides by $x$ gives $x = 5$ and loses the solution $x = 0$. Never divide by a variable that could be zero — instead move everything to one side and factor: $x^2 - 5x = 0 \Rightarrow x(x-5)=0$, giving $x = 0$ and $x = 5$.

Mistake 4: Mishandling the square root. $\sqrt{b^2 - 4ac}$ applies to the whole discriminant, not the terms separately — $\sqrt{9 + 40} = \sqrt{49} = 7$, not $\sqrt{9} + \sqrt{40}$.

Comparison: Which Method When?

Situation	Best method
Coefficients are small integers that factor cleanly	Factoring
You need the vertex or are deriving a formula	Completing the square
Nothing factors, or you want a guaranteed answer	Quadratic formula
You only need how many solutions	Discriminant alone

Practice Questions

Recall. Write the quadratic formula and name the discriminant within it.

Understanding. Explain why a quadratic can have two solutions but a linear equation has only one. (Hint: think about how many times each graph can cross the x-axis.)

Application. A rectangular garden has area 40 m² and its length is 3 m more than its width. Find the dimensions. (Set width = $w$; then $w(w+3) = 40 \Rightarrow w^2 + 3w - 40 = 0 \Rightarrow (w+8)(w-5)=0$. Width must be positive, so $w = 5$ m and length $= 8$ m.)

Analysis. Without solving, determine how many real solutions $3x^2 + 2x + 1 = 0$ has, and explain what that means geometrically. (Discriminant $= 2^2 - 4(3)(1) = 4 - 12 = -8 < 0$, so no real solutions — the parabola never touches the x-axis.)

FAQ

Why does $a$ have to be non-zero? If $a = 0$, the $x^2$ term vanishes and the equation becomes linear ($bx + c = 0$) — it's no longer quadratic.

What are the solutions when the discriminant is negative? There are no real solutions, but there are two complex solutions involving $i = \sqrt{-1}$. You'll meet these when you study complex numbers.

Is the quadratic formula always the best choice? It always works, but factoring is faster when the numbers are friendly. Learn to recognize easy factoring so you don't over-rely on the formula.

What's the difference between roots, solutions, and zeros? For a quadratic they're the same thing: the values of $x$ that make the expression equal zero — equivalently, where the graph crosses the x-axis.

How do I remember completing the square? Halve the coefficient of $x$, square it, and add it in. The "halve and square" step is the whole trick.

Quick Revision

• Standard form: $ax^2 + bx + c = 0$, $a \neq 0$.
• Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ — always use both signs.
• Discriminant $b^2 - 4ac$: positive → 2 roots, zero → 1 root, negative → 0 real roots.
• Vertex at $x = -\dfrac{b}{2a}$; parabola opens up if $a>0$, down if $a<0$.
• Never divide by a variable — factor instead, or you'll lose a solution.

Related Topics

Prerequisites: Linear Equations, factoring, and the laws of exponents.

Related: functions and graphing — a quadratic is the graph of a degree-2 function.

Next: polynomial equations of higher degree, and Calculus, where the vertex of a parabola is found instantly by setting the derivative to zero.