Quadratic Inequalities Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

Move every term to one side to compare with zero
Find the roots, which cut the line into sign intervals
Read a positive quadratic outside the roots when $a > 0$
Test a point in each interval to confirm the sign
Open or close the endpoints to match a strict or inclusive symbol
Handle a negative discriminant as all reals or no solution

What a quadratic inequality asks

A quadratic inequality in one variable is any of

\begin{aligned} ax^2 + bx + c &> 0, &\qquad ax^2 + bx + c &< 0, \\ ax^2 + bx + c &\ge 0, &\qquad ax^2 + bx + c &\le 0, \end{aligned}

with $a \ne 0$ . If the inequality does not already have $0$ on one side, your first move is to make it so. You do that by shifting every term to the same side, exactly as you would tidy an equation before solving. Once one side is $0$ , the whole problem becomes a question about the sign of the quadratic.

To see that, name the quadratic $y = ax^2 + bx + c$ , the very parabola from the previous lessons. Then $ax^2 + bx + c > 0$ is asking “for which $x$ is $y > 0$ ,” which means “where is the parabola above the x-axis.” In the same way $ax^2 + bx + c < 0$ asks where the parabola is below the axis. We are solving for the single variable $x$ , so every answer is a set of numbers on the number line, not a region of the plane.

The parabola y = x squared minus 2x minus 3 factors as (x + 1)(x minus 3), so it crosses the x-axis at the roots x = -1 and x = 3. Because it opens upward, the curve lies above the axis (where y is positive) outside the roots and dips below the axis (where y is negative) between them. The dashed lines mark the roots, where the sign switches.

Find the roots, then read the sign

Since the whole game is the sign of the quadratic, start where the sign can change. A quadratic changes sign only by passing through $0$ , which happens exactly at its roots. So solve the equation $ax^2 + bx + c = 0$ first, by factoring when it splits nicely or by the quadratic formula when it does not. The real roots are the x-intercepts of the parabola, and they cut the number line into intervals. Inside any one interval the quadratic never reaches $0$ , so it cannot flip sign there; it holds one sign all the way across.

There are two ways to decide which sign, and they always agree.

The first way reads the sign off the shape of the parabola. If $a > 0$ the parabola opens upward, so it climbs above the axis on both arms and dips below only in the middle. With two real roots on the line, such a parabola is positive outside the roots and negative between them. If $a < 0$ it opens downward and the picture is upside down, so the signs reverse.

The second way is a test point. Pick any convenient number inside an interval, substitute it into the quadratic, and note the sign of the result. That one value settles the sign for the entire interval, because the sign is constant there. Here is why both methods must give the same answer.

Why the sign is constant between two distinct roots, and flips at each one#

Suppose the quadratic $ax^2 + bx + c$ has two different real roots, and call them $r$ and $s$ with $r < s$ . Finding those roots is the factoring you already know, so the quadratic can be rewritten as $a(x - r)(x - s)$ . Its value at any $x$ is therefore a product of three things: the constant $a$ , the factor $x - r$ , and the factor $x - s$ . The sign of a product depends only on the signs of its parts, so track those.

Take a number to the right of both roots, $x > s$ . Then $x - r$ and $x - s$ are both positive, their product is positive, and the quadratic has the same sign as $a$ . Now slide left, between the roots, so $r < x < s$ . The factor $x - s$ has turned negative while $x - r$ is still positive, so their product is now negative, and the quadratic has the opposite sign from $a$ . Slide left once more, past both roots, so $x < r$ . Both factors are negative, their product is positive again, and the quadratic matches the sign of $a$ once more.

So crossing a single root flips exactly one factor, which flips the sign of the whole product, and nothing changes sign in between. That is why the sign stays constant on each interval and reverses at each of these two distinct roots. That is also why an upward parabola ( $a > 0$ ) is positive outside the roots and negative between them. It is the same story the picture tells: the curve can only move from one side of the x-axis to the other by passing through it. The curve passes through only at the roots. A repeated root brings the same factor back twice instead of two different factors, so it behaves differently, as the discriminant section later shows.

Take the parabola from the figure, $y = x^2 - 2x - 3 = (x + 1)(x - 3)$ , with roots $-1$ and $3$ . The table below runs the test-point method on all three intervals and matches it against the shape.

Interval	Test value	$(x+1)(x-3)$	Sign of $y$
$x < -1$	$x = -2$	$(-1)(-5) = 5$	positive
$-1 < x < 3$	$x = 0$	$(1)(-3) = -3$	negative
$x > 3$	$x = 4$	$(5)(1) = 5$	positive

Positive on the two outer intervals and negative in the middle, exactly as an upward parabola should be. With the sign chart in hand, answering any inequality on this quadratic is just a matter of keeping the intervals whose sign matches.

Worked example 1 Solve $x^2 - 2x - 3 > 0$

One side is already $0$ , so factor to find the roots:

x^2 - 2x - 3 = (x + 1)(x - 3), \qquad x = -1 \text{ or } x = 3.

The parabola opens upward ( $a = 1 > 0$ ), so it is positive outside the roots and negative between them. The inequality wants where the quadratic is greater than $0$ , that is, where the curve is above the axis, which is the two outer intervals. Because the symbol is strict, the roots themselves (where the value is exactly $0$ ) are excluded:

x < -1 \quad \text{or} \quad x > 3.

On a number line this is two open circles at $-1$ and $3$ with the shading running outward. Written with the interval notation from the linear-inequality chapter, it is $(-\infty, -1) \cup (3, \infty)$ , where the cup $\cup$ joins the two separate rays.

The solution of x squared minus 2x minus 3 greater than zero is x below -1 or x above 3. Open circles at the roots -1 and 3 show they are excluded because the inequality is strict, and the two shaded rays run outward, away from the gap between the roots.

Change the symbol and only the endpoints change. The inclusive inequality $x^2 - 2x - 3 \ge 0$ keeps the same two outer intervals but now includes the roots, where the value is $0$ . Its circles therefore fill in, and the answer is $x \le -1$ or $x \ge 3$ . Flip to $x^2 - 2x - 3 \le 0$ and you want where the curve is on or below the axis, the single middle interval with its endpoints, giving $-1 \le x \le 3$ .

Check your understanding

Solve the inequality $(x - 2)(x - 5) < 0$ .

2 < x < 5

x < 2

x > 5

2 \le x \le 5

-5 < x < -2

When the leading coefficient is negative

If $a < 0$ the parabola opens downward, so the sign pattern flips: the curve is now below the axis outside the roots and above it between them. You can read that straight off the shape, or you can sidestep it entirely by multiplying both sides of the inequality by $-1$ . Multiplying by $-1$ turns that negative leading coefficient positive and, by the rule from linear inequalities, reverses the symbol. Either route works, and they land on the same solution.

Worked example 2 Solve $-x^2 + 2x + 3 \ge 0$

The leading coefficient is negative, and there are two clean ways to handle that.

The first is to clear the negative. Multiply both sides by $-1$ , which flips the inequality symbol:

-x^2 + 2x + 3 \ge 0 \quad\Longrightarrow\quad x^2 - 2x - 3 \le 0.

Now it is an upward parabola again. Factoring gives $(x + 1)(x - 3) \le 0$ with roots $-1$ and $3$ , and an upward parabola is negative between its roots. Including the endpoints, where the value is $0$ , the solution is $-1 \le x \le 3$ .

The second way reads the sign straight off the downward parabola. With $a = -1 < 0$ the graph of $y = -x^2 + 2x + 3 = -(x + 1)(x - 3)$ opens downward, the mirror image of the upward case: above the axis between the roots, below it outside them. Asking for $y \ge 0$ again selects the middle interval including its endpoints, $-1 \le x \le 3$ , the same answer as the first route.

Flipping the sign of the leading coefficient reflects the picture. The downward parabola y = negative x squared plus 2x plus 3 equals negative (x + 1)(x minus 3) has the same roots -1 and 3, but now it rises above the axis (y positive) between the roots and falls below (y negative) outside them, the exact reverse of the upward case.

Check your understanding

Solve $-x^2 + 3x + 10 \ge 0$ .

-2 \le x \le 5

x \le -2

x \ge 5

-2 < x < 5

x < -2

x > 5

Rearrange first, and roots from the quadratic formula

Two practical wrinkles come up constantly. First, many inequalities arrive with terms on both sides, so you must gather everything onto one side before you can factor. Second, plenty of quadratics do not factor over the whole numbers, and there the quadratic formula supplies the roots. Neither changes the method; they only change how you reach the roots.

Worked example 3 Solve $x^2 + 5 \le 6x$

One side is not yet $0$ , so move everything to the left first. Subtract $6x$ from both sides:

x^2 + 5 \le 6x \quad\Longrightarrow\quad x^2 - 6x + 5 \le 0.

Now factor to find the roots:

x^2 - 6x + 5 = (x - 1)(x - 5), \qquad x = 1 \text{ or } x = 5.

The parabola opens upward, so it is negative between the roots, and the inclusive $\le$ keeps the endpoints. The solution is $1 \le x \le 5$ . Notice the trap this avoids: you cannot solve $x^2 + 5 \le 6x$ by dividing by $x$ , which would risk losing solutions and hide the sign. Moving everything to one side is what makes the sign analysis possible.

Worked example 4 Solve $x^2 - 4x + 1 > 0$

This one does not factor with whole numbers, so find the roots with the quadratic formula, using $a = 1$ , $b = -4$ , and $c = 1$ :

x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}.

So the roots are $2 - \sqrt{3} \approx 0.27$ and $2 + \sqrt{3} \approx 3.73$ . The parabola opens upward, so it is positive outside the roots, and the strict symbol excludes the roots themselves:

x < 2 - \sqrt{3} \quad \text{or} \quad x > 2 + \sqrt{3}.

A quick test confirms the choice of intervals. At $x = 0$ , which is outside the roots, the quadratic is $0 - 0 + 1 = 1 > 0$ ; at $x = 2$ , which is between them, it is $4 - 8 + 1 = -3 < 0$ . The positive values are indeed on the outside.

When the discriminant changes the story

The method so far assumed two separate roots. The discriminant $b^2 - 4ac$ tells you when that assumption holds and when it breaks down. A positive discriminant is exactly the case you have been solving all along: two distinct real roots, and the three-interval method from earlier applies directly. A negative discriminant means the parabola never touches the axis at all, so it keeps one sign for every $x$ , and the answer is either all real numbers or no solution, depending on the direction of the inequality and the sign of $a$ . A zero discriminant means the parabola touches the axis once, at a repeated root, and keeps the same sign everywhere except that single point; there the answer depends closely on the symbol, and it can be all real numbers, all real numbers except one point, no solution, or that one point alone. The worked example below covers the negative and zero cases.

Worked example 5 Two inequalities with no second root

Not every quadratic has two roots, and the discriminant tells you which case you are in.

First take $x^2 + 2x + 5 > 0$ . Its discriminant is

b^2 - 4ac = (2)^2 - 4(1)(5) = 4 - 20 = -16 < 0,

so there are no real roots. The parabola opens upward and never reaches the axis, so it floats entirely above it and is positive for every $x$ . The inequality $> 0$ is therefore true for all real numbers. Had the problem asked for $x^2 + 2x + 5 < 0$ , the answer would be the opposite extreme, no solution, since the curve is never below the axis.

Now take the repeated-root case $x^2 - 6x + 9 > 0$ . Here the quadratic is a perfect square,

x^2 - 6x + 9 = (x - 3)^2,

whose only root is $x = 3$ , where the upward parabola just touches the axis. The factor $x - 3$ changes sign as $x$ passes through $3$ , the same way it would for a distinct root, but squaring erases that change: a square is never negative, so it stays positive on both sides of $3$ and reaches $0$ only exactly at $x = 3$ . That is why the sign does not flip here. Everywhere else the square is positive, so $(x - 3)^2 > 0$ holds for every $x$ except $x = 3$ , giving $x < 3$ or $x > 3$ . Watch how the symbol changes this last case: $(x - 3)^2 \ge 0$ is true for all real numbers (a square is never negative). By contrast, $(x - 3)^2 \le 0$ holds only at the single point $x = 3$ , and $(x - 3)^2 < 0$ has no solution at all.

Check your understanding

Which describes the solution of $x^2 + 1 < 0$ ?

A no solution B all real numbers C

x < -1

-1 < x < 1

Worked example 6 When the variable is a real-world quantity

Not every quadratic inequality is about numbers on an abstract line. Suppose a garden is fenced into a rectangle whose length is $6$ meters more than its width $w$ , and its area must be at least $40$ square meters. That gives $w(w + 6) \ge 40$ , or after moving everything to one side, $w^2 + 6w - 40 \ge 0$ , which factors as $(w + 10)(w - 4) \ge 0$ with roots $w = -10$ and $w = 4$ .

The upward parabola is positive outside the roots, so the algebra alone gives $w \le -10$ or $w \ge 4$ . But $w$ is a width, and a width cannot be zero or negative. That rules out the whole ray $w \le -10$ : it solves the inequality, but it cannot describe a real garden. Only $w \ge 4$ meters remains, so the garden must be at least $4$ meters wide.

The method has not changed. What is new is a second check, after the algebra, against what the variable is actually allowed to represent. An algebraically correct piece of the answer can still be physically impossible.

Common mistakes

Choosing the wrong interval. For an upward parabola, $> 0$ is satisfied outside the roots and $< 0$ between them. Swapping these is the most common error. A single test point settles which is which every time.
Not comparing to zero first. Solve $x^2 < 6x$ by moving everything to one side, $x^2 - 6x < 0$ , then factoring. Dividing both sides by $x$ can lose solutions and destroys the sign picture.
Mismatched endpoints. A strict $<$ or $>$ excludes the roots (open circles); an inclusive $\le$ or $\ge$ includes them (closed circles). Match the circle to the final symbol.
Ignoring a negative leading coefficient. When $a < 0$ the parabola opens downward, so the “outside versus between” pattern reverses. Multiply through by $-1$ and flip the symbol to return to an upward parabola if that is clearer.
Assuming there are always two roots. If the discriminant is negative the quadratic keeps one sign everywhere, so the answer is all real numbers or no solution. A zero discriminant gives a repeated root where the curve only touches the axis.
Collapsing two rays into one band. An outside-the-roots answer is two separate pieces, $x < r$ or $x > s$ , joined by “or”. Writing it as $s < x < r$ names an empty, backwards interval and is meaningless.

Key takeaways

To solve a quadratic inequality, move every term to one side so it compares to $0$ , then find the roots of $ax^2 + bx + c$ by factoring or the quadratic formula.
The roots are the x-intercepts of the parabola $y = ax^2 + bx + c$ ; they cut the number line into intervals on which the quadratic holds a single sign.
For an upward parabola ( $a > 0$ ) the quadratic is positive outside the roots and negative between them; a downward parabola ( $a < 0$ ) reverses this. A test point in each interval confirms the sign.
Keep the intervals the inequality asks for: $> 0$ where the curve is above the axis, $< 0$ where it is below. A strict symbol excludes the roots (open circles); an inclusive symbol includes them (closed circles).
If the discriminant is negative there are no real roots, so the answer is all real numbers or no solution. A repeated root touches the axis at one point.
Present the answer as an inequality and on a number line. An outside-the-roots solution is two rays joined by “or”.

Quadratic Inequalities Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

What a quadratic inequality asks

Find the roots, then read the sign

When the leading coefficient is negative

Rearrange first, and roots from the quadratic formula

When the discriminant changes the story

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice