Quadratic Inequalities: Sign Charts & Parabolas

Definition and Standard Form

A square beats a line far out, but not in the middle.

Compare

x^2

with

3x + 10

. Far to the right the square wins easily, and far to the left too, because squaring makes negatives positive. Near zero the line is ahead. So

x^2 < 3x + 10

holds somewhere in the middle — but guessing values will never say exactly where it starts and stops.

x	x²	3x + 10	x² smaller?
−10	100	−20	no
0	0	10	yes
10	100	40	no

Working it out takes three moves.

Move everything to one side, so the question becomes a sign:

x^2 - 3x - 10 < 0

. Factor it:

(x + 2)(x - 5) < 0

. A product is negative only when its two factors have opposite signs, and each factor changes sign at one point —

x + 2

at

-2

,

x - 5

at

5

.

piece of the line	x + 2	x − 5	(x + 2)(x − 5)
left of −2	−	−	+
between −2 and 5	+	−	− : the answer
right of 5	+	+	+

Only the middle piece gives a negative product, so the answer is

-2 < x < 5

. Check one value from inside and one from outside: at

x = 0

,

0 < 10

is true; at

x = 6

,

36 < 28

is false. At

x = 5

both sides are

25

, which is not "smaller".

Before and after standard form

The section's x² < 3x + 10, with each side drawn as its own graph. The parabola dips below the line between their crossings at x = −2 and x = 5, so that stretch is the solution. Moving everything to one side gives x² − 3x − 10 < 0, whose roots are the same −2 and 5: rewriting in standard form changes how the question looks, not its answer.

Any inequality between two expressions can be read as one graph above or below another.

Nothing in that answer was chosen; each part was forced.

The borders

-2

and

5

were forced by the equation

x^2 - 3x - 10 = 0

: they are its two roots, the only places the expression can change sign. Which side to keep was forced by the leading coefficient:

a = 1

is positive, so the parabola opens upward, sits above the axis outside its roots and dips below only between them. Two roots and a sign — that is all the arithmetic was ever finding.

The dots are the roots

-2

and

5

. Between them the upward parabola is below the axis: the "−" row of the table.

The definition

A quadratic inequality in one variable is an inequality that can be written in the form below — or with

<

,

\leq

or

\geq

— where

a

,

b

,

c

are real constants and

a \neq 0

.

Definition and Standard Form

ax^2 + bx + c > 0

Its equation counterpart: quadratic equations.

A quadratic inequality is a statement about the sign of a quadratic, and its answer is a set.

It is not a number and not a function. It asks where a quadratic polynomial is positive or negative, and its answer collects those values — for our example, the interval

(-2, 5)

. Depending on the roots and on

a

, the answer can be one interval, two rays, a single point, every real number or none at all.

The bar is the middle piece of the table; the open dots are the two roots, left out because

<

is strict.

Two things carry over to every quadratic inequality, and two were peculiar to this one.

What carries over: the borders are always the roots of the matching equation, and the sign of

a

always decides between inside and outside. What was peculiar: the example had two different roots and an upward parabola. Turn the parabola over and the same roots give the opposite answer. With one repeated root, or none —

x^2 + 4 < 0

has no solution at all — there are fewer borders or none, and the answer is a single point, everything or nothing.

The same roots

-2

and

5

with opposite

a

: an upward parabola keeps the inside, a downward one keeps the two outside rays.

The rest of the page sorts quadratic inequalities by the discriminant — two roots, one root, no roots — then gives the sign chart method, the connection to the parabola, inequalities that reduce to quadratics, and a summary of every case.

The Role of the Discriminant

The discriminant

\Delta = b^2 - 4ac

determines how many real roots the quadratic has, and therefore how many boundary points appear on the number line. This settles the structure of the solution before any interval testing begins.

When

\Delta > 0

, two distinct real roots

r_1 < r_2

exist. They divide the number line into three intervals:

(-\infty, r_1)

,

(r_1, r_2)

, and

(r_2, \infty)

. The sign of the quadratic alternates across these intervals, and the solution is either the middle interval, the two outer intervals, or some combination depending on the inequality direction and the sign of

a

.

When

\Delta = 0

, one repeated root

r

exists. The quadratic factors as

a(x - r)^2

, which is a perfect square scaled by

a

. The expression equals zero at

x = r

and has the sign of

a

everywhere else. The number line is effectively split into two intervals with identical sign, separated by a single point of zero.

When

\Delta < 0

, no real roots exist. The quadratic never touches zero and maintains one sign throughout the entire real line — the sign of

a

. The inequality is either satisfied everywhere or nowhere, with no intervals to analyze.

Discriminant Δ = b² − 4ac	# of real roots	# of intervals	Sign behavior of the quadratic
Δ > 0	2 distinct (r₁ < r₂)	3	sign alternates: outer regions opposite to inner
Δ = 0	1 repeated (r)	2	sign of a everywhere; equals 0 only at r
Δ < 0	none	1 (whole line)	sign of a everywhere; never zero

A quadratic can change sign only at a real root, and the discriminant counts those roots before any testing.

Solving When the Discriminant Is Positive

When

\Delta > 0

, the two roots

r_1 < r_2

create three intervals. The sign of the quadratic in these intervals follows a fixed pattern determined by the leading coefficient

a

.

RULE

If

a > 0

, the parabola opens upward. The quadratic is positive on the outer intervals

(-\infty, r_1)

and

(r_2, \infty)

, and negative on the inner interval

(r_1, r_2)

.

The expression dips below zero between the roots and rises above zero outside them.

Pitfall: Multiplying by -1 without flipping the sign

Turning

-x^2 + x + 6 > 0

into

x^2 - x - 6 > 0

keeps the sign and gives

x < -2

or

x > 3

. Multiplying by

-1

flips it:

x^2 - x - 6 < 0

, so

-2 < x < 3

.

If

a < 0

, the parabola opens downward. The signs are reversed: the quadratic is negative on the outer intervals and positive on the inner interval

(r_1, r_2)

. The expression rises above zero only between the roots.

Example

To solve

x^2 - 5x + 4 > 0

, find the roots of

x^2 - 5x + 4 = 0

. Factoring gives

(x - 1)(x - 4) = 0

, so

r_1 = 1

and

r_2 = 4

. Since

a = 1 > 0

, the expression is positive outside the roots. The solution is

(-\infty, 1) \cup (4, \infty)

.

To solve

x^2 - 5x + 4 \leq 0

instead, select the interval where the expression is non-positive:

[1, 4]

. The endpoints are included because the inequality is non-strict and the expression equals zero at the roots.

Δ > 0: the parabola dips between its two roots

With two roots the parabola crosses the axis twice and lies below it exactly between the crossings, so ax² + bx + c < 0 holds on (−2, 3) and the reverse inequality holds on the two outer rays. The leading coefficient decides which of the two patterns applies. Flip a and watch the shading invert on the inequality visual explorer.

The two-root case is the only one where the answer is a genuine interval with two endpoints.

Solving When the Discriminant Is Zero

When

\Delta = 0

, the quadratic has a single repeated root

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

, and the expression factors as

a(x - r)^2

.

RULE

The square

(x - r)^2

is non-negative for all real

x

and equals zero only at

x = r

. The sign of the entire expression is therefore determined by

a

alone, except at the single point

x = r

where the expression is zero.

If

a > 0

: the expression is positive everywhere except at

x = r

, where it is zero. The inequality

ax^2 + bx + c > 0

is satisfied by all

x \neq r

, and the solution is

(-\infty, r) \cup (r, \infty)

. The non-strict version

ax^2 + bx + c \geq 0

is satisfied by every real number:

(-\infty, \infty)

.

Pitfall: Missing the root in a non-strict inequality

Reading

(x - 2)^2 \leq 0

as "a square is never negative, so no solution" drops the equality. Since

x^2 - 4x + 4 = (x - 2)^2

equals

0

at

x = 2

, the solution is the single point

x = 2

.

If

a < 0

: the expression is negative everywhere except at

x = r

, where it is zero. The inequality

ax^2 + bx + c < 0

is satisfied by all

x \neq r

, and the non-strict version

ax^2 + bx + c \leq 0

holds for all real

x

. The inequality

ax^2 + bx + c > 0

has no solution — a negative times a non-negative square is never positive.

Example

Consider

x^2 - 6x + 9 > 0

. The discriminant is

36 - 36 = 0

, and the expression factors as

(x - 3)^2

. This is positive for all

x \neq 3

and zero at

x = 3

. The solution is

(-\infty, 3) \cup (3, \infty)

, the entire real line minus one point.

Fact: Why the square wins on area

A rectangle with perimeter

20

and one side

x

has area

x(10 - x)

, and

25 - x(10 - x) = (x - 5)^2 \geq 0

. So no rectangle with that perimeter beats the

5 \times 5

square, and only the square reaches

25

.

The same perfect square settles questions that do not look like inequalities at first. When an expression can be rewritten as a constant minus a square,

k - (x - r)^2

, it can never be larger than

k

, because the square is never negative, and it equals

k

only at

x = r

. The claim that

k

is the largest value is then just the inequality

(x - r)^2 \geq 0

, the

\Delta = 0

case of this section with

a > 0

.

A single root that never changes the sign

With Δ = 0 the quadratic is a perfect square, here (x − 2)². It touches the axis at x = 2 and is positive everywhere else. So (x − 2)² > 0 holds for every x except 2 (the open circle), ≥ 0 holds for every real number, < 0 never holds, and ≤ 0 holds only at x = 2.

The repeated root is the only point where a strict inequality can fail.

Solving When the Discriminant Is Negative

RULE

When

\Delta < 0

, the quadratic has no real roots. It never equals zero and never changes sign. The sign of the expression for every real

x

is the sign of the leading coefficient

a

.

If

a > 0

: the expression is positive everywhere. The inequality

ax^2 + bx + c > 0

(or

\geq 0

) is satisfied by every real number, and the solution is

(-\infty, \infty)

. The inequality

ax^2 + bx + c < 0

(or

\leq 0

) has no solution.

If

a < 0

: the expression is negative everywhere. The inequality

ax^2 + bx + c < 0

(or

\leq 0

) holds for all

x

, and

ax^2 + bx + c > 0

(or

\geq 0

) has no solution.

Pitfall: Reading no roots as no solution

Concluding from

\Delta = 4 - 20 = -16

that

x^2 + 2x + 5 > 0

has no solution answers the equation, not the inequality. With

a = 1 > 0

the expression is always positive, so every real

x

is a solution.

Example

The equation

x^2 + x + 1 = 0

has

\Delta = 1 - 4 = -3 < 0

, and since

a = 1 > 0

, the expression

x^2 + x + 1

is positive for every real

x

. The inequality

x^2 + x + 1 > 0

is universally true. The inequality

x^2 + x + 1 < 0

is universally false.

No sign chart is needed in this case. The absence of real roots means no critical points exist, so there is only one interval — the entire number line — and its sign is determined by

a

.

Fact: The discriminant proves Cauchy–Schwarz

For any numbers

p, q, r, s

, the quadratic

(pt + r)^2 + (qt + s)^2

in

t

is never negative, so its discriminant is at most

0

. Written out, that is

(pr + qs)^2 \leq (p^2 + q^2)(r^2 + s^2)

, the inequality behind

|\cos\theta| \leq 1

for the angle between two vectors.

The reasoning also runs backwards. If a quadratic with

a > 0

is known to be non-negative for every real

x

, it cannot have two distinct real roots, because it would be negative between them, so its discriminant must satisfy

\Delta \leq 0

. A statement about the sign of an expression at every point then becomes a single inequality between its coefficients, with no

x

left in it.

Δ < 0: x² + 4 never reaches the axis

The parabola floats above the axis, never touching it, so x² + 4 is positive for every real x: the inequality x² + 4 < 0 has no solution at all, and x² + 4 > 0 has every real number. With no roots there are no boundaries and the sign is constant. See the empty solution set on the inequality visual explorer.

A negative discriminant makes the inequality trivial one way and impossible the other.

Sign Chart Method

The sign chart provides a systematic alternative to reasoning from the parabola's shape. It works by tracking the sign of each linear factor across the intervals created by the roots.

For the inequality

(x - 1)(x - 4) > 0

, the roots are

r_1 = 1

and

r_2 = 4

, creating three intervals. In the interval

(-\infty, 1)

, both factors

x - 1

and

x - 4

are negative, so their product is positive. In

(1, 4)

, the factor

x - 1

is positive while

x - 4

is negative, so the product is negative. In

(4, \infty)

, both factors are positive, so the product is positive. The sign chart summarizes this:

Sign Chart Method

\begin{array}{c|ccc} & x < 1 & 1 < x < 4 & x > 4 \\ \hline (x-1) & - & + & + \\ (x-4) & - & - & + \\ \hline \text{product} & + & - & + \end{array}

Pitfall: Setting each factor positive

Solving

(x + 2)(x - 3) > 0

as

x + 2 > 0

or

x - 3 > 0

gives

x > -2

, which includes

x = 0

, where the product is

-6

. The product is positive when both factors share a sign:

x < -2

or

x > 3

.

The solution to the

> 0

inequality consists of the intervals where the product is positive:

(-\infty, 1) \cup (4, \infty)

.

The same method as a list of steps, with the example beside each one.

DownloadPNG2×PDF

Step

(x - 1)(x - 4) > 0

1

Find the rootsThey are the boundary points of the chart

r_1 = 1

,

r_2 = 4

2

List the intervals the roots create

(-\infty, 1)

,

(1, 4)

,

(4, \infty)

3

Find the sign of each linear factor in each interval

x - 1

:

-

,

+

,

+

and

x - 4

:

-

,

-

,

+

4

Multiply the signs to get the sign of the product

product:

+

,

-

,

+

5

Keep the intervals whose sign matches the inequalityFor

> 0

, the intervals where the product is positive

(-\infty, 1) \cup (4, \infty)

The sign chart method, step by step·/algebra/inequalities/quadraticLearn Math Class

When the quadratic does not factor neatly, the roots from the quadratic formula serve the same purpose. The sign chart does not require integer or rational roots — any pair of real numbers works as boundary points. The method fails only when no real roots exist (

\Delta < 0

), in which case the sign is constant and no chart is needed.

To solve your own quadratic inequality with each step shown, use the quadratic inequality solver.

Connection to the Parabola

The inequality

ax^2 + bx + c > 0

asks a geometric question: where does the parabola

y = ax^2 + bx + c

lie above the

x

-axis? The inequality

ax^2 + bx + c < 0

asks where it lies below.

When

a > 0

, the parabola opens upward. It dips below the axis between its roots (if they exist) and rises above the axis outside them. The

> 0

inequality selects the outer regions; the

< 0

inequality selects the inner region.

When

a < 0

, the parabola opens downward. It rises above the axis between its roots and falls below outside. The

> 0

inequality selects the inner region; the

< 0

inequality selects the outer regions.

This geometric reading makes the sign chart unnecessary for many students — once the roots and the direction of opening are identified, the solution is visible from the shape of the curve. The sign chart is the algebraic encoding of the same information, and either approach yields identical results.

The vertex of the parabola, located at

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

, is the point of maximum depth below the axis (when

a > 0

and the parabola crosses the axis) or maximum height above it (when

a < 0

). Its position does not directly affect the solution set of the inequality, but it determines how far the parabola penetrates into the region of the opposite sign.

Leading coefficient	Parabola opens	Above x-axis (expression > 0)	Below x-axis (expression < 0)
a > 0	upward	outside the roots: (−∞, r₁) ∪ (r₂, ∞)	between the roots: (r₁, r₂)
a < 0	downward	between the roots: (r₁, r₂)	outside the roots: (−∞, r₁) ∪ (r₂, ∞)

A quadratic inequality is a question about a parabola: its roots and the way it opens decide the answer.

Reducible Inequalities

Certain inequalities of degree higher than two can be reduced to quadratic inequalities through substitution, following the same strategy used for reducible equations.

Example

The bi-quadratic inequality

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

becomes

u^2 - 5u + 4 > 0

with the substitution

u = x^2

. The quadratic factors as

(u - 1)(u - 4) > 0

, which is satisfied when

u < 1

or

u > 4

. Back-substituting:

x^2 < 1

gives

-1 < x < 1

, and

x^2 > 4

gives

x < -2

or

x > 2

. The complete solution is

(-\infty, -2) \cup (-1, 1) \cup (2, \infty)

.

Pitfall: Taking only the positive square root

Turning

x^2 > 9

into

x > 3

loses half the answer:

x = -4

also works, since

(-4)^2 = 16 > 9

. The solution is

x < -3

or

x > 3

.

The back-substitution step requires care. The condition

u = x^2

forces

u \geq 0

, so any solution for

u

that is negative produces no real values of

x

. Additionally, each positive value of

u

splits into two symmetric values

x = \pm\sqrt{u}

, potentially doubling the number of boundary points on the

x

-axis.

The substitution pattern extends to any inequality of the form

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

. Setting

u = f(x)

reduces it to a quadratic inequality in

u

, and the solution in

u

is translated back through the function

f

— subject to whatever domain and range constraints

f

imposes.

A quartic inequality solved as a quadratic

The section's x⁴ − 5x² + 4 > 0. With u = x² it becomes (u − 1)(u − 4) > 0, true for u < 1 or u > 4. Back in x, u < 1 is −1 < x < 1 and u > 4 is x < −2 or x > 2: exactly the three stretches where the curve is above the axis, with the four boundaries ±1 and ±2 excluded.

Each positive value of u yields two symmetric boundaries in x.

Summary of Quadratic Inequality Cases

All quadratic inequalities reduce to one of three structural cases determined by the discriminant

\Delta

. The table below collects each case with the resulting sign pattern of the expression and the solutions to the strict

> 0

and

< 0

inequalities, assuming

a > 0

. For

a < 0

, the

> 0

and

< 0

solutions swap. For non-strict inequalities (

\geq 0

or

\leq 0

), include the roots where the expression equals zero.

Discriminant	Roots	Sign pattern across the line (a > 0)	Solution to expression > 0	Solution to expression < 0
Δ > 0	r₁ < r₂	+ − + (across r₁, r₂)	(−∞, r₁) ∪ (r₂, ∞)	(r₁, r₂)
Δ = 0	r (repeated)	+ everywhere; equals 0 at r	all x ≠ r	∅ (empty)
Δ < 0	none	+ everywhere	(−∞, ∞)	∅ (empty)

Common Pitfalls

Five mistakes account for most wrong answers to quadratic inequalities. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

DownloadPNG2×PDF

1Multiplying by -1 without flipping the sign

The leading coefficient is negative, and the inequality is multiplied by

-1

to make it positive, which makes factoring easier. Every term changes sign, but the inequality sign is left as it was.

Start

-x^2 + x + 6 > 0

Wrong turn

Error here

\implies x^2 - x - 6 > 0

\implies (x + 2)(x - 3) > 0

\implies x < -2 \text{ or } x > 3

Right path

Multiplying both sides by a negative number reverses the inequality.

\implies x^2 - x - 6 < 0

\implies (x + 2)(x - 3) < 0

\implies -2 < x < 3

Why it happensMultiplying by

-1

feels like tidying up the expression, and the inequality sign is forgotten as part of the same operation.

How to checkTest one value.

x = 0

gives

-0 + 0 + 6 = 6 > 0

, so

0

must be in the answer; it lies in

-2 < x < 3

but not in the wrong answer. The shape agrees: with

a = -1 < 0

the parabola opens downward and is positive between its roots.

2Missing the root in a non-strict inequality

When

\Delta = 0

, the expression is a multiple of a perfect square, and the inequality includes equality. The square is treated as if it could only be positive or negative, and the point where it equals zero is lost.

Start

x^2 - 4x + 4 \leq 0

\implies (x - 2)^2 \leq 0

Wrong turn

Error here

And since a square is never negative, the inequality is declared to have no solution.

Right path

(x - 2)^2 \leq 0

means

(x - 2)^2 < 0

or

(x - 2)^2 = 0

. The first never happens; the second happens at

x = 2

. The solution is the single point

x = 2

.

Why it happensThe familiar fact that a square is never negative answers the strict question

< 0

, and the "or equal" part of

\leq

gets dropped.

How to checkSubstitute the repeated root.

x = 2

gives

4 - 8 + 4 = 0

, and

0 \leq 0

is true, so the solution set cannot be empty.

3Reading no roots as no solution

The discriminant is negative, so the corresponding equation has no real roots, and that verdict is carried over to the inequality.

Start

x^2 + 2x + 5 > 0, \quad \Delta = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0

Wrong turn

Error here

So the inequality is declared to have no solution.

Right path

With

\Delta < 0

the expression never changes sign, and its sign is the sign of

a

. Here

a = 1 > 0

, so

x^2 + 2x + 5 > 0

for every real

x

. The solution is

(-\infty, \infty)

.

Why it happensFor equations, a negative discriminant does mean no real solutions, and the same verdict is applied out of habit. An inequality asks about the sign of the expression, not about where it equals zero.

How to checkSubstitute any value.

x = 0

gives

5 > 0

, so the solution set is not empty. Completing the square shows why it holds everywhere:

x^2 + 2x + 5 = (x + 1)^2 + 4 \geq 4

.

4Setting each factor positive

After factoring, the inequality

(x - r_1)(x - r_2) > 0

is solved like an equation, by applying the condition to each factor separately.

Start

(x + 2)(x - 3) > 0

Wrong turn

Error here

\implies x + 2 > 0 \text{ or } x - 3 > 0

\implies x > -2

Right path

A product of two factors is positive when both factors have the same sign. Both are positive when

x > 3

; both are negative when

x < -2

. The solution is

(-\infty, -2) \cup (3, \infty)

.

Why it happensThe zero-product property, a product is zero when one factor is zero, is carried over to inequalities. There is no such rule for "greater than zero": a product is positive only when its factors share a sign, which is a condition on both factors at once.

How to checkTest one value from the wrong answer.

x = 0

gives

(0 + 2)(0 - 3) = -6 < 0

. Values from the right answer,

x = -3

and

x = 4

, give

(-1)(-6) = 6

and

(6)(1) = 6

, both positive.

5Taking only the positive square root

After back-substitution, an inequality of the form

x^2 > p

or

x^2 < p

with

p > 0

remains, and the square root is taken as if both sides were positive numbers.

Start

x^2 > 9

Wrong turn

Error here

\implies x > 3

Right path

x^2 > 9

means

|x| > 3

, so

x < -3

or

x > 3

. In the same way,

x^2 < 9

means

|x| < 3

, that is

-3 < x < 3

.

Why it happensTaking the square root of both sides is the step used for equations like

x^2 = 9

, where the

\pm

is easy to remember. In an inequality the negative side becomes a separate interval, and it is left out.

How to checkTest a negative value.

x = -4

gives

(-4)^2 = 16 > 9

, so it belongs to the answer, yet

-4 > 3

is false. The same test catches the

<

case:

x < 3

would include

x = -5

, where

25 < 9

is false.

Common pitfalls in quadratic inequalities·/algebra/inequalities/quadraticLearn Math Class

Three of the slips carry a habit from equations over to inequalities: taking a single square root, setting each factor positive on its own, and reading no roots as no solution. The other two lose the direction or the equality part of the sign. A single test value catches every one of them, as the checks show, and a quick sketch of the parabola, with its roots and the way it opens, explains the right answer.

Quadratic Inequalities FAQ

Why is the solution to $x^2 - 6x + 9 > 0$ everything except one point?

+

Because the expression factors as

(x - 3)^2

, which is positive everywhere except at

x = 3

, where it equals zero. Zero fails a strict inequality, so that single point drops out and the answer is

(-\infty, 3) \cup (3, \infty)

. Switch to

\geq 0

and the point returns, making the solution every real number.Read more →

Can a quadratic inequality have no solution at all?

+

Yes, whenever the direction contradicts the sign the expression always has. With

\Delta < 0

and

a > 0

the expression is positive everywhere, so

ax^2 + bx + c < 0

is satisfied by nothing. The same happens at a double root: with

a < 0

, the expression is never positive, so

> 0

has no solution.Read more →

If a quadratic has no real roots, do you still need a sign chart?

+

No. Real roots are what carve the number line into intervals, so without them there is a single interval, the whole line, and one sign to determine. That sign is the sign of the leading coefficient

a

. Evaluate the discriminant, check

a

, and the answer is either all reals or nothing at all.Read more →

How can you tell that a quadratic is never negative?

+

Check two numbers:

ax^2 + bx + c \geq 0

for every real

x

exactly when

a > 0

and

\Delta = b^2 - 4ac \leq 0

. With two distinct real roots the expression would dip below zero between them, so

\Delta

cannot be positive, and with

a < 0

it is negative far from the vertex. For example,

x^2 + 2x + 5

has

a = 1

and

\Delta = -16

, and indeed it equals

(x + 1)^2 + 4

.Read more →

Why does a square have the largest area of any rectangle with the same perimeter?

+

Because the shortfall is a perfect square. With perimeter

20

and one side

x

, the other side is

10 - x

, and

25 - x(10 - x) = x^2 - 10x + 25 = (x - 5)^2 \geq 0

. So no such rectangle has area above

25

, and only

x = 5

, the

5 \times 5

square, reaches it. Any perimeter

P

works the same way, with

\frac{P}{4}

in place of

5

.Read more →

Can you solve $(x + 2)(x - 3) > 0$ by making each factor positive?

+

No. A product is positive when both factors share a sign, not when either one is positive. Setting

x + 2 > 0

or

x - 3 > 0

gives

x > -2

, which includes

x = 0

, where the product is

(2)(-3) = -6

. Both factors are positive for

x > 3

and both negative for

x < -2

, so the solution is

(-\infty, -2) \cup (3, \infty)

.Read more →

Quadratic Inequalities

Where the Parabola Meets the Number Line

Key Terms

Core Concepts

From Other Categories

Definition and Standard Form

The Role of the Discriminant

Solving When the Discriminant Is Positive

Solving When the Discriminant Is Zero

Solving When the Discriminant Is Negative

Sign Chart Method

Connection to the Parabola

Reducible Inequalities

Summary of Quadratic Inequality Cases

Common Pitfalls

Quadratic Inequalities FAQ

Why is the solution to $x^2 - 6x + 9 > 0$ everything except one point?

Can a quadratic inequality have no solution at all?

If a quadratic has no real roots, do you still need a sign chart?

How can you tell that a quadratic is never negative?

Why does a square have the largest area of any rectangle with the same perimeter?

Can you solve $(x + 2)(x - 3) > 0$ by making each factor positive?

Quadratic Inequalities

Where the Parabola Meets the Number Line

Key Terms

Core Concepts

From Other Categories

Definition and Standard Form

The Role of the Discriminant

Solving When the Discriminant Is Positive

Solving When the Discriminant Is Zero

Solving When the Discriminant Is Negative

Sign Chart Method

Connection to the Parabola

Reducible Inequalities

Summary of Quadratic Inequality Cases

Common Pitfalls

Quadratic Inequalities FAQ

Why is the solution to x2−6x+9>0x^2 - 6x + 9 > 0x2−6x+9>0 everything except one point?

Can a quadratic inequality have no solution at all?

If a quadratic has no real roots, do you still need a sign chart?

How can you tell that a quadratic is never negative?

Why does a square have the largest area of any rectangle with the same perimeter?

Can you solve (x+2)(x−3)>0(x + 2)(x - 3) > 0(x+2)(x−3)>0 by making each factor positive?

Why is the solution to $x^2 - 6x + 9 > 0$ everything except one point?

Can you solve $(x + 2)(x - 3) > 0$ by making each factor positive?