Quadratic Equations and the Discriminant

⚡ Key Takeaways

A quadratic equation is any equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$ . Its solutions are given by the quadratic formula, and the sign of the discriminant $\Delta = b^2 - 4ac$ tells you instantly how many real solutions exist (two, one, or none) without solving. Used everywhere in algebra, physics, engineering, and economics. Studied from Babylonian times (~2000 BCE) through al-Khwārizmī (9th c.) to its modern algebraic form.

💡 The Core Insight

Every quadratic can be reshaped into a perfect square plus a constant (completing the square). Once in that form, solving is just taking a square root. The number under that square root is the discriminant — and since you cannot take the real square root of a negative number, the sign of the discriminant alone decides whether real solutions exist, how many, and whether they are equal. The quadratic formula is simply completing the square done once, in general, for all quadratics at once.

DefinitionQuadratic Equation

A quadratic equation in one variable $x$ is any equation equivalent to

1

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

Derivation hint: This is the general degree-2 polynomial set equal to zero; the condition $a\neq 0$ guarantees the $x^2$ term survives, otherwise it would be linear. where:

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

A value of $x$ satisfying the equation is called a root or solution.

DefinitionThe Discriminant

The discriminant of $ax^2 + bx + c = 0$ is the quantity

2

\quad \Delta = b^2 - 4ac

Derivation hint: It is precisely the expression that appears under the square root sign in the quadratic formula; its name comes from "discriminating" between root types.

TheoremQuadratic Formula

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

3

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

Derivation hint: Obtained by completing the square on the general quadratic; the $\pm$ encodes the two square roots. Moreover, the discriminant classifies the roots:

Discriminant	Real roots	Nature
$\Delta > 0$	two distinct	two real, unequal
$\Delta = 0$	one repeated	one real, "double root"
$\Delta < 0$	none	two complex conjugates

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

ProofDeriving the Quadratic Formula by Completing the Square

We solve $ax^2 + bx + c = 0$ in general.

Step 1. Since $a \neq 0$ , divide through by $a$ :

5

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

Derivation hint: Dividing by the leading coefficient makes the $x^2$ term monic so we can complete the square cleanly.

Step 2. Move the constant to the right:

6

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

Derivation hint: Isolate the variable terms so the left side can be built into a perfect square.

Step 3. Add $\left(\dfrac{b}{2a}\right)^2$ to both sides to complete the square. Half of the coefficient of $x$ is $\frac{b}{2a}$ :

7

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

Derivation hint: $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$ is exactly the term needed to make a perfect-square trinomial.

Step 4. The left side is now a perfect square; combine the right side over a common denominator $4a^2$ :

8

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

Derivation hint: $-\frac{c}{a} = -\frac{4ac}{4a^2}$ , so the right side becomes $\frac{b^2-4ac}{4a^2} = \frac{\Delta}{4a^2}$ .

Step 5. Take the square root of both sides (the $\pm$ captures both roots):

9

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

Derivation hint: $\sqrt{4a^2} = 2|a|$ , and the $\pm$ absorbs the sign so we may write $2a$ in the denominator.

Step 6. Solve for $x$ :

10

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

Derivation hint: Subtract $\frac{b}{2a}$ from both sides and combine the two fractions over $2a$ . This is equation [3]. The expression under the root, $b^2-4ac = \Delta$ , controls whether the root is real (when $\Delta \geq 0$ ) or complex (when $\Delta < 0$ ). $\blacksquare$

∎

mathematician · Muḥammad ibn Mūsā al-Khwārizmī | c. 780 – c. 850 CE

The Persian polymath whose treatise al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wa-l-muqābala ("The Compendious Book on Calculation by Completion and Balancing") gave us the word algebra (from al-jabr, "restoration/completion"). He classified quadratics into six standard types — because negative numbers were not yet accepted — and solved each geometrically by literally completing a square. He gave systematic, general procedures rather than one-off tricks, laying the foundation for algebra as a discipline.

🧠 Intuition

Picture the parabola $y = ax^2 + bx + c$ . Solving the equation means asking: where does this curve cross the x-axis (where $y=0$ )?

If the parabola dips through the axis, it crosses at two points → two real roots → $\Delta > 0$ .
If it just touches the axis at its vertex, it crosses at one point → a repeated root → $\Delta = 0$ .
If it floats entirely above (or below) the axis, it never crosses → no real roots → $\Delta < 0$ .

The vertex sits at $x = -\frac{b}{2a}$ , and the $\pm\sqrt{\Delta}/2a$ in the formula measures how far left and right the crossings lie from the vertex. When $\Delta = 0$ , that spread shrinks to zero and the two crossings merge into the vertex itself.

Figure

A parabola and its discriminant: three cases of where

y=ax^2+bx+c

meets the x-axis

ExampleTwo Real Roots — A Projectile

A ball is thrown upward; its height in metres after $t$ seconds is $h(t) = -5t^2 + 20t + 25$ . When does it hit the ground ( $h = 0$ )?

Given: $a = -5$ , $b = 20$ , $c = 25$ .

Step 1 — Discriminant:

11

\quad \Delta = b^2 - 4ac = 20^2 - 4(-5)(25) = 400 + 500 = 900

Derivation hint: Substitute directly into $\Delta = b^2-4ac$ ; watch the double negative $-4(-5)(25)=+500$ . Since $\Delta = 900 > 0$ , there are two real roots, and $\sqrt{900}=30$ .

Step 2 — Quadratic formula:

12

\quad t = \frac{-20 \pm 30}{2(-5)} = \frac{-20 \pm 30}{-10}

Derivation hint: Plug $-b=-20$ , $\sqrt\Delta=30$ , $2a=-10$ into formula [3]. This gives $t = \frac{10}{-10} = -1$ or $t = \frac{-50}{-10} = 5$ .

Step 3 — Interpret: Time cannot be negative, so we discard $t=-1$ . > Answer: The ball hits the ground at $\boxed{t = 5 \text{ s}}$ .

ExampleOne Repeated Root and No Real Roots

(a) Solve $x^2 - 6x + 9 = 0$ .

Given: $a=1,\ b=-6,\ c=9$ .

13

\quad \Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0

Derivation hint: $\Delta=0$ signals a perfect-square trinomial: $x^2-6x+9=(x-3)^2$ . Since $\Delta = 0$ , there is one repeated root:

14

\quad x = \frac{-(-6)\pm\sqrt{0}}{2(1)} = \frac{6}{2} = 3

Derivation hint: With $\sqrt\Delta=0$ the $\pm$ collapses, leaving a single value $-b/2a$ . > Answer: $\boxed{x = 3}$ (a double root).

(b) Solve $x^2 + x + 1 = 0$ .

Given: $a=1,\ b=1,\ c=1$ .

15

\quad \Delta = 1^2 - 4(1)(1) = 1 - 4 = -3

Derivation hint: Negative discriminant means no real solutions; the roots are complex conjugates.

16

\quad x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

Derivation hint: $\sqrt{-3} = \sqrt{3}\,\sqrt{-1} = i\sqrt3$ , where $i^2=-1$ . > Answer: $\boxed{x = -\tfrac{1}{2} \pm \tfrac{\sqrt3}{2}\,i}$ — two complex conjugate roots, no real solutions.

⚠️ Frequent Errors

1. Forgetting $a \neq 0$ . If $a=0$ the equation is linear, not quadratic, and the formula divides by zero. 2. Sign of $-b$ . The formula uses $-b$ . If $b=-6$ , then $-b=+6$ . Dropping this flips your answers. 3. The $\pm$ is not optional. A quadratic generically has two roots; using only $+$ loses one. 4. "No real roots" ≠ "no roots." When $\Delta<0$ there are still two complex roots. 5. Discriminant sign vs. magnitude. Only the sign of $\Delta$ classifies the roots; its size doesn't matter for counting. 6. $-4ac$ with negative $c$ . $-4ac$ where $a>0, c<0$ becomes positive, increasing $\Delta$ — a common slip with double negatives.

CounterexampleWhy "

x^2 = -4 \Rightarrow x = -2

" Is Wrong

A student writes $x^2 + 4 = 0 \Rightarrow x^2 = -4 \Rightarrow x = -2$ . Check: $(-2)^2 = +4 \neq -4$ . The equation $x^2+4=0$ has $\Delta = 0^2 - 4(1)(4) = -16 < 0$ and therefore no real roots at all. The correct solutions are the complex numbers $x = \pm 2i$ . No real number, positive or negative, has a negative square — squaring always yields a non-negative result.

connection

The discriminant connects many neighbouring ideas:

Completing the Square is the engine behind the quadratic formula; the formula is completing the square done generally.
Parabolas and Graphs of Quadratic Functions give the geometric meaning: roots are x-intercepts, $\Delta$ counts crossings, and the vertex sits at $x=-b/2a$ .
Complex Numbers supply the "missing" roots when $\Delta<0$ , making the Fundamental Theorem of Algebra hold: a degree- $n$ polynomial has exactly $n$ roots counted with multiplicity.
Vieta's Formulas relate roots to coefficients without solving, equation [4].
Factoring Polynomials is the shortcut when roots are rational; a perfect-square $\Delta$ signals a clean factorization.
The discriminant generalizes to the Discriminant of Higher-Degree Polynomials, detecting repeated roots in cubics and beyond.

🔬 Where this appears

⏳ History

- c. 2000 BCE — Babylonians: Clay tablets show systematic methods for solving problems equivalent to quadratics (e.g., finding a number given its sum and product), using procedures like completing the square — though purely numerically and without negative roots.

- c. 300 BCE — Euclid: Elements Book II solves quadratic problems geometrically via "application of areas," providing rigorous constructions.

- c. 250 CE — Diophantus: In Arithmetica, works with quadratic equations symbolically but accepts only positive rational roots.

- c. 628 CE — Brahmagupta: The Indian mathematician gives a near-complete verbal rule for solving quadratics, explicitly using negative numbers and zero, and recognizing two roots.

- c. 820 CE — al-Khwārizmī: Systematizes the solution into six canonical types and proves the method by geometric completion of the square; coins the term al-jabr (algebra).

- 12th c. — Bhāskara II: Extends quadratic methods and recognizes both positive and negative roots in his Līlāvatī and Bījagaṇita.

- 16th c. — Cardano, Bombelli: While solving cubics, Bombelli (1572) is forced to take square roots of negatives, introducing what would become complex numbers — the home of roots when

\Delta<0

.

- 1637 — René Descartes: La Géométrie introduces modern symbolic notation (

a,b,c

for constants,

x

for the unknown) and connects equations to curves.

- 18th–19th c.: The general algebraic quadratic formula and the term discriminant are standardized; Gauss (1799) proves the Fundamental Theorem of Algebra, guaranteeing every quadratic has exactly two roots over the complex numbers.

open-problem · Beyond the Quadratic

The quadratic formula expresses roots using only $+,-,\times,\div$ and square roots — radicals. This raises a deep question: can every polynomial equation be solved by radicals?

Cubics (degree 3) and quartics (degree 4) can: Cardano (1545) and Ferrari gave general radical formulas, with their own discriminants distinguishing root types.
Quintics (degree 5) and higher generally cannot: the Abel–Ruffini theorem (1824) and Galois theory (1830s) prove no general formula in radicals exists. The discriminant survives as a tool but no longer guarantees a closed-form solution.

Even at the quadratic level there are subtleties: over different number systems (modular arithmetic, finite fields, the p-adics) the discriminant's meaning shifts — whether $\Delta$ is a "square" in that system replaces the simple sign test. The classification " $\Delta>0,=0,<0$ " is special to the real numbers, where order and squares interact.

Quadratic Equations and the Discriminant

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also