Quadratic Equations

A quadratic equation can be rearranged into $ax^{2} + bx + c = 0$ with $a \ne 0$ . It can have two, one or no real solutions.

There is no single best method for every quadratic. Factoring is efficient when the expression factors cleanly; square-root methods are efficient when a squared expression is isolated; completing the square and the quadratic formula handle more general cases.

On this page (8 sections)

Factoring and the zero-product property

Over the real numbers, a product is zero exactly when at least one factor is zero. First rearrange the equation so one side is zero.

Solve $x^2-5x+6=0$

$(x-2)(x-3)=0$
Factor the quadratic.
$x-2=0\quad\text{or}\quad x-3=0$
Apply the zero-product property.
$x=2\quad\text{or}\quad x=3$
Solve the two linear equations.

Both answers check: $2^2-5(2)+6=0$ and $3^2-5(3)+6=0$ . A product equal to $40$ does not allow you to set its factors equal to zero.

Square-root method

If a square is isolated, use both possible signs when its value is positive.

(x-4)^2=9

x-4=\pm3

x=4\pm3\quad\Longrightarrow\quad x=1\ \text{or}\ x=7

The principal square root $\sqrt9$ is $3$ ; the equation $u^2=9$ has the two solutions $u=3$ and $u=-3$ . These are different statements.

One real solution or none

If $(x-4)^2=0$ , only $x=4$ works. If $(x-4)^2=-9$ , there is no real solution because a real square cannot be negative.

Roots and x-intercepts

A real solution of $x^2-5x+6=0$ is an $x$ -coordinate where $y=x^2-5x+6$ meets the horizontal axis. This graph crosses at $(2,0)$ and $(3,0)$ .

A real root is an input where the output is zero

$y=x^2-5x+6$

The roots are

x=2

and

x=3

Diagram explained

The parabola crosses the horizontal axis at the two marked points. Its vertex lies between them at (5/2, −1/4), just below the axis.

Exact reference values for this graph
Input	Output
$0$	$6$
$2$	$0$
$5/2$	$-1/4$
$3$	$0$
$5$	$6$

An exact factorization establishes the roots. A plotted curve is a representation of the result, not a substitute for checking the equation.

Choosing a solution method

Choosing a quadratic-solving method
Structure	Useful first method
product equals zero	factoring
$(x – h)^{2} = k$	square-root method
general quadratic	completing the square or quadratic formula
graphing/technology context	graph to estimate, then verify algebraically

Context can remove algebraic roots

A rectangle has width $x$ centimetres, length $x+3$ centimetres and area $40$ square centimetres. The model requires $x>0$ .

x(x+3)=40\quad\Longleftrightarrow\quad x^2+3x-40=0

(x+8)(x-5)=0\quad\Longrightarrow\quad x=-8\ \text{or}\ x=5

Both numbers solve the equation, but only $x=5$ satisfies the physical domain. The dimensions are 5 cm by 8 cm, and $5\cdot8=40$ square centimetres checks the area. The rejected negative root is a valid algebraic root, not a calculation error.

Practice questions

Solve $x^{2} – 5x + 6 = 0$ .
Solve $(x – 4)^{2} = 9$ .
Why might an algebraic root be rejected in an application?

Check your answers

$(x-2)(x-3)=0$ , so $x=2$ or $x=3$ .
$x-4=\pm3$ , so $x=1$ or $x=7$ .
An equation can have roots outside a model’s domain. A negative value, for example, cannot represent the length of this rectangle.

Sources and further reading

OpenStax: Quadratic functions and equations. Definitions, conventions and related examples.
OpenStax Elementary Algebra 2e. Language of algebra, equations, inequalities and factoring.
OpenStax Intermediate Algebra 2e. Quadratic equations, factoring and algebraic techniques.
Wolfram Language: Algebraic Transformations. Symbolic simplification, expansion and factoring reference.