What is a quadratic equation?

A quadratic equation has the form ax² + bx + c = 0 where a ≠ 0. The variable appears to the second power as its highest degree. Every quadratic can be brought into standard form by rearranging terms.

How do you solve quadratic equations by factoring?

Factor the quadratic into two binomials: (px + q)(rx + s) = 0. By the zero-product property, set each factor to zero and solve the resulting linear equations. This works when the quadratic factors neatly over rationals.

What is completing the square?

Completing the square transforms ax² + bx + c = 0 into (x + k)² = d form. Add (b/2a)² to both sides to create a perfect square trinomial, then take the square root. This method works universally for all quadratics.

What is the quadratic formula?

The quadratic formula x = (-b ± √(b²-4ac))/2a gives both solutions directly from the coefficients. It's derived from completing the square on the general equation and works for any quadratic regardless of factorability.

What does the discriminant tell you?

The discriminant Δ = b² - 4ac determines the nature of solutions. If Δ > 0: two distinct real roots. If Δ = 0: one repeated (double) root. If Δ < 0: no real roots, but two complex conjugate roots.

What are Vieta's formulas for quadratics?

For ax² + bx + c = 0 with roots x₁ and x₂: the sum x₁ + x₂ = -b/a and the product x₁·x₂ = c/a. They verify solutions, construct equations from known roots, and can solve simple quadratics by inspection — find two numbers with the right sum and product.

What are reducible equations?

Equations like ax⁴ + bx² + c = 0 (bi-quadratic) can be reduced to quadratics by substituting u = x². Solve for u, then back-substitute to find x = ±√u for each positive u value.

How do quadratic equations relate to parabolas?

Solutions of ax² + bx + c = 0 are the x-intercepts of y = ax² + bx + c. The discriminant determines whether the parabola crosses the axis twice (Δ > 0), touches once (Δ = 0), or never touches (Δ < 0).

Quadratic Equations

Q: What are Vieta's formulas for quadratics?

For ax² + bx + c = 0 with roots x₁ and x₂: the sum x₁ + x₂ = -b/a and the product x₁·x₂ = c/a. They verify solutions, construct equations from known roots, and can solve simple quadratics by inspection — find two numbers with the right sum and product.

Q: What are reducible equations?

Equations like ax⁴ + bx² + c = 0 (bi-quadratic) can be reduced to quadratics by substituting u = x². Solve for u, then back-substitute to find x = ±√u for each positive u value.

Q: How do quadratic equations relate to parabolas?

Solutions of ax² + bx + c = 0 are the x-intercepts of y = ax² + bx + c. The discriminant determines whether the parabola crosses the axis twice (Δ > 0), touches once (Δ = 0), or never touches (Δ < 0).

Key Terms

Definition and Standard Form

Solving by Factoring

Solving by Completing the Square

The Quadratic Formula

The Discriminant

Vieta's Formulas

Reducible Equations

Connection to Graphing

Summary of Solving Methods

Second-Degree Equations and the Threshold of Multiplicity

When the unknown is squared, equations acquire a new dimension of complexity. A quadratic equation can produce two solutions, one solution, or no real solutions at all — and a single quantity built from its coefficients determines which case applies before any solving begins. Several distinct methods handle the solving itself, each suited to different situations but all ultimately equivalent. Quadratic equations sit at the boundary between elementary algebra and the richer structure of polynomial equations, and the tools developed here — the discriminant, factoring, completing the square, the quadratic formula, and Vieta's formulas — recur throughout mathematics.

Key Terms

Equation Structure

Standard Form—

ax^2 + bx + c = 0

with

a \neq 0

Degree of an Equation— quadratic equations have degree

2

Coefficient— the values

a

b

c

that determine the equation's behavior

Solutions

Discriminant—

\Delta = b^2 - 4ac

, determines the number and type of solutions

Solution Set— may contain two, one, or zero real values

Extraneous Solution— may arise from non-reversible steps during solving

Formulas Used on This Page

Quadratic Formula—

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

Discriminant—

\Delta = b^2 - 4ac

Square Root Property—

x^2 = p \implies x = \pm\sqrt{p}

Completing the Square—

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

Vieta's Formulas (Quadratic)—

x_1 + x_2 = -\frac{b}{a}

x_1 \cdot x_2 = \frac{c}{a}

See All Algebra Definitions →

See All Algebra Formulas →

Definition and Standard Form

A quadratic equation in one variable is any equation that can be written as

ax^2 + bx + c = 0

where

a

b

, and

c

are real constants and

a \neq 0

. The requirement

a \neq 0

is what makes the equation genuinely quadratic — if

a

were zero, the

x^2

term would vanish and the equation would reduce to a linear equation.

The word "quadratic" derives from the Latin word for square, reflecting the presence of

x^2

as the highest-degree term. The coefficient

a

is the leading coefficient,

b

is the linear coefficient, and

c

is the constant term. Every quadratic equation can be brought into standard form by rearranging: moving all terms to one side and ordering by descending powers of

x

.

The equation

x^2 = 5x - 6

, for instance, is not in standard form but is still quadratic. Subtracting

5x

and adding

6

to both sides yields

x^2 - 5x + 6 = 0

, with

a = 1

b = -5

, and

c = 6

. Recognition is the first step — any equation reducible to the form above, regardless of how it is initially presented, is quadratic and amenable to the methods on this page.

Solving by Factoring

When the quadratic expression

ax^2 + bx + c

can be decomposed into a product of two first-degree factors, the equation reduces to a pair of linear equations through the zero-product property: if a product of two factors equals zero, at least one of the factors must be zero.

Consider

x^2 - 5x + 6 = 0

. The left side factors as

(x - 2)(x - 3) = 0

. Setting each factor equal to zero gives

x - 2 = 0

x - 3 = 0

, so

x = 2

x = 3

. Both values satisfy the original equation.

When the leading coefficient is not

1

, factoring requires more care. The equation

6x^2 + x - 2 = 0

factors as

(2x - 1)(3x + 2) = 0

, giving

x = \frac{1}{2}

x = -\frac{2}{3}

. Finding these factors involves searching for two binomials whose product reconstructs the original trinomial — a process that amounts to finding two numbers whose product is

ac

and whose sum is

b

.

Factoring is efficient when it works, but it has a fundamental limitation: not every quadratic with rational coefficients factors into binomials with rational entries. The equation

x^2 - 2 = 0

has solutions

x = \pm\sqrt{2}

, which are irrational, so no factorization over the rationals exists. A method that handles every quadratic without exception requires a different approach.

Solving by Completing the Square

Completing the square transforms any quadratic equation into a form where the solution can be extracted by taking a square root. The method works unconditionally — it does not depend on whether the quadratic factors neatly.

Starting from

ax^2 + bx + c = 0

, divide every term by

a

to obtain

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

. Move the constant to the right side:

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

The left side is almost a perfect square trinomial. Adding

\left(\frac{b}{2a}\right)^2

to both sides completes it:

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

The left side factors as a perfect square:

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

Taking the square root of both sides and isolating

x

gives the two solutions. The expression under the square root,

b^2 - 4ac

, determines whether those solutions are real or complex.

As a concrete example, consider

x^2 + 6x + 2 = 0

. Move the constant:

x^2 + 6x = -2

. Half of

6

3

, and

3^2 = 9

. Add

9

to both sides:

x^2 + 6x + 9 = 7

. Factor:

(x + 3)^2 = 7

. Take roots:

x + 3 = \pm\sqrt{7}

. The solutions are

x = -3 + \sqrt{7}

and

x = -3 - \sqrt{7}

The Quadratic Formula

Applying the method of completing the square to the general equation

ax^2 + bx + c = 0

— with letters in place of specific numbers — produces a universal formula:

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

This is the quadratic formula. It accepts any real coefficients

a

b

c

(with

a \neq 0

) and returns the solutions directly. The

\pm

symbol indicates that two values are generated: one using the positive square root and one using the negative.

For the equation

2x^2 - 3x - 5 = 0

, the coefficients are

a = 2

b = -3

c = -5

. Substituting:

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}

The two solutions are

x = \frac{10}{4} = \frac{5}{2}

and

x = \frac{-4}{4} = -1

.

The formula is not a separate technique from completing the square — it is the result of completing the square once, in general form, so that the work never needs repeating. Every specific application of completing the square arrives at the same place the formula reaches in a single substitution.

The Discriminant

The expression

\Delta = b^2 - 4ac

appearing under the square root in the quadratic formula is called the discriminant. Its value determines the nature and number of solutions without requiring the full computation.

When

\Delta > 0

, the square root is a positive real number, and the formula produces two distinct real solutions. The equation

x^2 - 5x + 3 = 0

has

\Delta = 25 - 12 = 13 > 0

, so two different real roots exist.

When

\Delta = 0

, the square root vanishes, and the formula collapses to a single value

x = -\frac{b}{2a}

. The quadratic has a repeated root — a double root — and the polynomial factors as

a(x - r)^2

where

r = -\frac{b}{2a}

. The equation

x^2 - 6x + 9 = 0

has

\Delta = 36 - 36 = 0

and the single solution

x = 3

.

When

\Delta < 0

, no real number has a negative square, so the square root does not produce a real value. Over the real numbers, the equation has no solutions. Over the complex numbers, the formula yields two conjugate roots

x = \frac{-b \pm i\sqrt{|\Delta|}}{2a}

, where

i

is the imaginary unit. The equation

x^2 + x + 1 = 0

has

\Delta = 1 - 4 = -3 < 0

and the complex roots

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

.

The discriminant is a diagnostic tool: it classifies the equation's solution structure from the coefficients alone.

Discriminant Δ = b² − 4ac	# real roots	Type of roots	Example
Δ > 0	2	two distinct real roots	x² − 5x + 3 = 0 (Δ = 13)
Δ = 0	1	one repeated (double) real root	x² − 6x + 9 = 0 (Δ = 0)
Δ < 0	0	two complex conjugate roots	x² + x + 1 = 0 (Δ = −3)

Vieta's Formulas

The French mathematician François Viète established a direct relationship between the roots of a quadratic equation and its coefficients. For the equation

ax^2 + bx + c = 0

with roots

x_1

and

x_2

x_1 + x_2 = -\frac{b}{a}

x_1 \cdot x_2 = \frac{c}{a}

These identities follow from factoring the quadratic as

a(x - x_1)(x - x_2) = 0

and expanding:

a[x^2 - (x_1 + x_2)x + x_1 x_2] = 0

. Matching coefficients with

ax^2 + bx + c

produces both formulas.

Vieta's formulas serve three purposes. First, they verify solutions: after solving

x^2 - 7x + 10 = 0

and obtaining

x_1 = 2

x_2 = 5

, check that

2 + 5 = 7 = -\frac{-7}{1}

and

2 \cdot 5 = 10 = \frac{10}{1}

. Both match, confirming the answers.

Second, they construct equations from known roots. If a quadratic equation must have roots

x_1 = -3

and

x_2 = 4

, then

x_1 + x_2 = 1

and

x_1 \cdot x_2 = -12

, so the equation is

x^2 - x - 12 = 0

(taking

a = 1

). The roots determine the equation uniquely up to a constant multiple.

Third — and often overlooked — they can be used to solve quadratic equations directly when the roots are simple integers. Instead of factoring or applying the formula, the question becomes: which two numbers sum to

-\frac{b}{a}

and multiply to

\frac{c}{a}

? For

x^2 - 7x + 10 = 0

, this asks for two numbers summing to

7

and multiplying to

10

— namely

2

and

5

. The roots are read off by inspection. This approach is fastest when applicable but limited: it works only when the roots are obvious enough to spot, which restricts it to quadratics with integer or simple rational solutions. For irrational or complex roots, the formula remains necessary. In practice, Vieta's inspection is essentially a special case of factoring — finding the factorization

(x - x_1)(x - x_2)

by going through the roots rather than the trinomial — but it is fast enough on simple cases to deserve recognition as a method in its own right.

Quantity	In terms of roots	In terms of coefficients	Use
Sum	x₁ + x₂	−b ⁄ a	solve by inspection, verify, construct from roots
Product	x₁ · x₂	c ⁄ a	solve by inspection, verify, construct from roots

Reducible Equations

Certain equations of degree higher than two can be converted into quadratic equations through substitution. The key is recognizing that the equation's structure mirrors the quadratic pattern

au^2 + bu + c = 0

, with some expression playing the role of

u

.

The most common case is the bi-quadratic equation

ax^4 + bx^2 + c = 0

. Setting

u = x^2

transforms it into

au^2 + bu + c = 0

, a standard quadratic in

u

. Solving for

u

yields up to two values, and each positive value of

u

produces two values of

x

via

x = \pm\sqrt{u}

. A negative value of

u

contributes no real solutions for

x

.

For example,

x^4 - 5x^2 + 4 = 0

becomes

u^2 - 5u + 4 = 0

with

u = x^2

. Factoring gives

(u - 1)(u - 4) = 0

, so

u = 1

u = 4

. Back-substituting:

x^2 = 1

gives

x = \pm 1

, and

x^2 = 4

gives

x = \pm 2

. The original quartic has four real solutions.

The substitution pattern extends beyond powers of

x

. Any equation of the form

a[f(x)]^2 + b[f(x)] + c = 0

is quadratic in

f(x)

. Setting

u = f(x)

reduces it, and back-substitution after solving requires solving

f(x) = u_1

and

f(x) = u_2

separately. Extraneous solutions can arise during back-substitution, so every candidate must be checked in the original equation.

Original form	Substitute	Becomes	Recover x by
ax⁴ + bx² + c = 0 (bi-quadratic)	u = x²	au² + bu + c = 0	x = ±√u for each u ≥ 0
a[f(x)]² + b·f(x) + c = 0 (general)	u = f(x)	au² + bu + c = 0	solve f(x) = u₁ and f(x) = u₂; verify all candidates

Connection to Graphing

The solutions of

ax^2 + bx + c = 0

correspond to the points where the parabola

y = ax^2 + bx + c

meets the

x

-axis. At these points the output value is zero, which is precisely what the equation demands. The discriminant controls the geometry of this intersection.

When

\Delta > 0

, the parabola crosses the

x

-axis at two distinct points, and the equation has two real solutions. The horizontal distance between these crossing points is

\frac{\sqrt{\Delta}}{|a|}

, so a larger discriminant means the roots are farther apart.

When

\Delta = 0

, the parabola is tangent to the

x

-axis — it touches at exactly one point and turns back. The vertex of the parabola sits on the axis, and the double root

x = -\frac{b}{2a}

is the

x

-coordinate of that vertex.

When

\Delta < 0

, the parabola lies entirely above the

x

-axis (if

a > 0

) or entirely below it (if

a < 0

). It never reaches the axis, so no real intersection exists and the equation has no real solutions. The vertex is the closest the parabola comes to the axis, and the gap between them reflects how negative the discriminant is.

The sign of

a

determines the parabola's orientation — opening upward when

a > 0

and downward when

a < 0

— but it does not affect the number of solutions. That role belongs exclusively to the discriminant.

Discriminant	x-axis intersection	Geometric behavior
Δ > 0	two distinct points	parabola crosses the axis; gap between roots is √Δ ⁄ \|a\|
Δ = 0	one point (tangent)	vertex sits on the axis at x = −b ⁄ 2a
Δ < 0	none	parabola entirely above (a > 0) or below (a < 0) the axis

Summary of Solving Methods

Four methods cover every quadratic equation in practice. Two are universal — completing the square and the quadratic formula always work. Two depend on the equation's specific structure: factoring needs a clean rational decomposition, and Vieta's inspection needs roots simple enough to spot. The table below summarizes when to reach for each, whether it is universal, and what insight each one produces beyond the solution itself.

Method	When to use	Universal?	Notes
Factoring	when the quadratic factors over the rationals	✗	fastest when applicable; uses the zero-product property
Completing the square	always	✓	conceptual root; produces the quadratic formula in the general case
Quadratic formula	always	✓	direct substitution; surfaces the discriminant explicitly
Vieta's / inspection	when integer roots are visible from sum and product	✗	guess two numbers with sum −b ⁄ a and product c ⁄ a; only practical when roots are simple