Quadratic Equations

⚡ Key Takeaways

A quadratic equation is any equation reducible to $ax^2 + bx + c = 0$ with $a \neq 0$ . Its (at most two) solutions are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Quadratics govern projectile motion, optimization, areas, and countless physical and economic models. Known in geometric form to the Babylonians (~2000 BCE); fully algebraic by the Islamic Golden Age and Renaissance.

💡 The Core Insight

Every quadratic can be rewritten as a perfect square plus a constant — a process called completing the square. This single algebraic move transforms the messy general equation $ax^2 + bx + c = 0$ into something of the form $(\text{stuff})^2 = \text{number}$ , which is solved by taking a square root. The quadratic formula is nothing more than completing the square done once, abstractly, for all quadratics at once.

DefinitionQuadratic Equation

A quadratic equation in the variable $x$ is an equation that can be written in the standard form

1

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

Derivation hint: This is the definition; the condition $a\neq 0$ ensures the equation is genuinely degree two and not linear. where:

$a$ is the leading coefficient (real, nonzero),
$b$ is the linear coefficient (real),
$c$ is the constant term (real),
$x$ is the unknown (the variable to solve for).

The values of $x$ satisfying the equation are called its roots or solutions.

TheoremThe Quadratic Formula

The roots of $ax^2 + bx + c = 0$ (with $a \neq 0$ ) are exactly

2

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

Derivation hint: Obtained by completing the square on the standard form — see the proof below. The quantity under the radical,

3

\quad \Delta = b^2 - 4ac,

Derivation hint: It is the radicand of equation [2]; its sign determines whether the square root is real, zero, or imaginary. is called the discriminant. It classifies the roots:

$\Delta > 0$ : two distinct real roots,
$\Delta = 0$ : one repeated real root (a double root),
$\Delta < 0$ : two complex-conjugate roots.

TheoremVieta's Formulas

If $x_1$ and $x_2$ are the two roots of $ax^2 + bx + c = 0$ , then

4

\quad x_1 + x_2 = -\frac{b}{a}, \qquad x_1 x_2 = \frac{c}{a}

Derivation hint: Expand $a(x-x_1)(x-x_2)=ax^2 - a(x_1+x_2)x + ax_1x_2$ and match coefficients with $ax^2+bx+c$ .

ProofDeriving the Quadratic Formula by Completing the Square

Step 1 — Divide by the leading coefficient. Since $a \neq 0$ , divide [1] by $a$ :

5

\quad x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Derivation hint: Dividing by $a$ makes the leading coefficient 1, which is required before completing the square.

Step 2 — Move the constant to the right side.

6

\quad x^2 + \frac{b}{a}x = -\frac{c}{a}

Derivation hint: Subtract $c/a$ from both sides to isolate the variable terms.

Step 3 — Complete the square. Add the square of half the linear coefficient, $\left(\frac{b}{2a}\right)^2$ , to both sides:

7

\quad x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}

Derivation hint: Half of $b/a$ is $b/(2a)$ ; its square is $b^2/(4a^2)$ . Adding to both sides preserves equality.

Step 4 — Factor the left side as a perfect square and combine the right side over a common denominator:

8

\quad \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Derivation hint: The left side is the perfect-square trinomial $\left(x+\tfrac{b}{2a}\right)^2$ ; the right combines $-c/a = -4ac/4a^2$ with $b^2/4a^2$ .

Step 5 — Take square roots of both sides, remembering the $\pm$ :

9

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

Derivation hint: $\sqrt{4a^2} = 2|a|$ , but the $\pm$ absorbs the sign, so $2a$ is used.

Step 6 — Solve for $x$ . Subtract $\frac{b}{2a}$ :

10

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

Derivation hint: Combine the two fractions over the common denominator $2a$ to recover equation [2]. ∎

∎

mathematician · Muhammad ibn Musa al-Khwarizmi | c. 780–850 CE

The Persian scholar al-Khwarizmi, working in Baghdad's House of Wisdom, wrote Al-Kitab al-mukhtasar fi hisab al-jabr wal-muqabala ("The Compendious Book on Calculation by Completion and Balancing") around 820 CE. The word al-jabr — "restoration," meaning moving a subtracted term to the other side — became our word algebra. He classified quadratics into six types (lacking negative numbers and zero as coefficients) and solved each with geometric "completing the square" demonstrations. His Latinized name, Algoritmi, gave us the word algorithm.

🧠 Intuition

Picture the parabola $y = ax^2 + bx + c$ . Solving the equation $ax^2 + bx + c = 0$ means finding where this curve crosses the x-axis (where $y=0$ ).

If the parabola dips below the axis and comes back up, it pierces the axis at two points — two real roots ( $\Delta > 0$ ).
If the parabola just touches the axis at its vertex, the two roots have merged into one ( $\Delta = 0$ ).
If the parabola floats entirely above (or below) the axis, never touching it, there are no real crossings — the roots are complex ( $\Delta < 0$ ).

The vertex sits at $x = -\frac{b}{2a}$ , exactly midway between the two roots — which is precisely why the formula has $-\frac{b}{2a}$ as its center and $\frac{\sqrt{\Delta}}{2a}$ as the symmetric "spread" on either side.

Figure

A parabola crossing the x-axis at its two real roots; the vertex lies midway between them.

ExampleTwo Distinct Real Roots

Solve $2x^2 - 7x + 3 = 0$ .

Given: $a = 2$ , $b = -7$ , $c = 3$ .

Step 1 — Compute the discriminant:

\Delta = b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25.

Since $\Delta = 25 > 0$ , expect two distinct real roots.

Step 2 — Apply the quadratic formula:

x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}.

Step 3 — Evaluate both signs:

x = \frac{7+5}{4} = \frac{12}{4} = 3, \qquad x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}.

Check with Vieta: $x_1 + x_2 = 3 + \tfrac12 = \tfrac72 = -b/a = 7/2$ ✓ and $x_1 x_2 = 3 \cdot \tfrac12 = \tfrac32 = c/a$ ✓.

\boxed{x = 3 \quad \text{or} \quad x = \tfrac{1}{2}}

ExampleProjectile Motion — A Physical Quadratic

A ball is thrown upward from a height of 1.5 m with initial speed 12 m/s. When does it hit the ground? Use $g = 9.8\ \text{m/s}^2$ .

Given: height model $h(t) = -\tfrac{1}{2}g t^2 + v_0 t + h_0 = -4.9 t^2 + 12 t + 1.5$ (metres). Set $h = 0$ :

-4.9 t^2 + 12 t + 1.5 = 0.

Step 1 — Identify coefficients: $a = -4.9$ , $b = 12$ , $c = 1.5$ .

Step 2 — Discriminant:

\Delta = 12^2 - 4(-4.9)(1.5) = 144 + 29.4 = 173.4.

$\sqrt{173.4} \approx 13.168$ .

Step 3 — Formula:

t = \frac{-12 \pm 13.168}{2(-4.9)} = \frac{-12 \pm 13.168}{-9.8}.

t = \frac{-12 + 13.168}{-9.8} \approx -0.119\ \text{s} \quad(\text{rejected: negative time}),

t = \frac{-12 - 13.168}{-9.8} \approx 2.568\ \text{s}.

We keep the physically meaningful (positive) root.

\boxed{t \approx 2.57\ \text{s}}

⚠️ Common mistake

The condition $a \neq 0$ is not optional. If $a = 0$ , equation [1] collapses to the linear equation $bx + c = 0$ , and the quadratic formula divides by zero ( $2a = 0$ ). Always check that the highest-degree term truly exists before applying the formula.

Two further frequent errors:

Forgetting the $\pm$ — every nonzero discriminant gives two roots, not one.
Mishandling signs of $b$ . The formula uses $-b$ , so when $b = -7$ , the term becomes $-(-7) = +7$ . Substitute carefully.
Writing the denominator as $2a$ but applying it only to the radical, not to $-b$ . The entire numerator $-b \pm \sqrt{\Delta}$ is divided by $2a$ .

CounterexampleA "Quadratic" That Isn't

Consider $(x-1)^2 = x^2 - 4$ . It looks quadratic, but expanding gives

x^2 - 2x + 1 = x^2 - 4 \;\Longrightarrow\; -2x + 1 = -4 \;\Longrightarrow\; x = \tfrac{5}{2}.

The $x^2$ terms cancel, leaving a linear equation with a single solution. The lesson: classify an equation only after simplifying to standard form — apparent degree can be deceptive.

connection

Quadratic equations sit at a crossroads of high-school mathematics:

They are the degree-two case of Polynomial Equations; cubics and quartics have their own (far more complex) formulas, but degree-five and higher have none in general.
Their graphs are Parabolas, the simplest curved conic sections, so the algebra here directly informs the geometry of Conic Sections.
Solving $\Delta < 0$ cases requires Complex Numbers, where $\sqrt{-1} = i$ makes every quadratic solvable.
The technique of Completing the Square reappears in integration, in deriving the vertex form, and in Gaussian probability.
Vieta's Formulas generalize to relate the coefficients and roots of any polynomial.

🔬 Where this appears

⏳ History

- c. 2000 BCE — Babylon. Clay tablets (e.g., BM 13901) show scribes solving problems equivalent to quadratics by geometric "cut-and-paste" methods, effectively completing the square — though only positive solutions were sought.

- c. 300 BCE — Greece. Euclid's Elements (Book II) gives geometric constructions for quadratic problems via the "application of areas."

- c. 628 CE — India. Brahmagupta, in the Brahmasphutasiddhanta, states a rule essentially equivalent to the quadratic formula and was among the first to use negative numbers and zero.

- c. 820 CE — Baghdad. Al-Khwarizmi systematically classifies and solves quadratics, founding algebra as a discipline.

- c. 1100 CE — Spain/Persia. Omar Khayyam advances geometric solutions of cubics, building on the quadratic apparatus.

- 1145 CE — Europe. Robert of Chester translates al-Khwarizmi's Algebra into Latin, transmitting the methods to the West.

- 1545 CE — Italy. Gerolamo Cardano's Ars Magna publishes general solutions for cubics and quartics and confronts square roots of negatives, foreshadowing complex numbers.

- 1637 CE — France. René Descartes introduces modern symbolic notation (

x

, exponents) in La Géométrie, putting quadratics in essentially their present form.

- 1799 CE — Germany. Carl Friedrich Gauss proves the Fundamental Theorem of Algebra, guaranteeing that every quadratic has exactly two complex roots.

open-problem

While quadratic equations are completely solved by a formula, the broader question of solvability by radicals is deep and historically explosive. The Abel–Ruffini theorem (1824) proves that the general polynomial of degree five or higher has no solution formula in radicals — no quintic analogue of the quadratic formula can exist. Galois theory (1830s) explains why: the symmetry group of a polynomial's roots must be "solvable," and the symmetric group $S_5$ is not. Thus the friendly quadratic formula is, in a precise sense, near the boundary of what algebra can deliver explicitly. Open and active areas include efficient numerical root-finding for high-degree and multivariate polynomial systems (computational algebraic geometry), where no closed forms exist and stability of computed roots remains subtle.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also