Quadratic Inequalities

Intuition

You can already find where a quadratic is zero. Factoring and the quadratic formula both answer the question $ax^2 + bx + c = 0$ , the inputs where the parabola meets the $x$ -axis. But most real questions are not about the single instant a quantity is zero. A ball is above ten feet for a stretch of time, not at one instant, and a business turns a profit over a range of prices, not at one price. Those are questions about where a quadratic is positive or negative, and that is a quadratic inequality. This chapter treats a quadratic as one object wearing three forms. Of those forms, it was the factored form $a(x - r_1)(x - r_2)$ that was advertised as the one handing you the zeros and the sign. This lesson cashes in the second half of that promise. The whole thing rests on one structural fact, that a quadratic can change sign only at a root. Once you have that fact, the answer is no longer a number but a solution set. That is exactly the kind of object the systems chapter taught you to name, draw, and combine directly.

Learning goals

Argue that a quadratic changes sign only at a root
Split the line at the roots and read each sign
Record the answer in interval notation, joining with a union
Open or close endpoints for strict or non-strict symbols
Let $\Delta$ decide the shape of the answer

A quadratic changes sign only at a root

Everything in this lesson follows from a single proposition, so it is worth stating it plainly and proving it before using it. A quadratic with two real roots factors, by the last few lessons, as $f(x) = a(x - r_1)(x - r_2)$ . The claim concerns the sign of such a product across the number line. The product is positive on some stretches and negative on others, but it can only switch between the two at $r_1$ or $r_2$ .

Why a quadratic changes sign only at a root#

Write the quadratic in factored form, $f(x) = a(x - r_1)(x - r_2)$ with $a \ne 0$ . Its sign at any input is the product of exactly three signs: the sign of the constant $a$ , together with the sign of each linear factor. So control the factors and you control the sign of $f$ .

Look at one factor, $x - r_1$ . It equals zero exactly at $x = r_1$ . For every $x > r_1$ it is positive, and for every $x < r_1$ it is negative. So a single linear factor is positive on one side of its own root, negative on the other, and changes sign nowhere else. The same is true of $x - r_2$ at $r_2$ .

Now take any interval of the number line that contains neither root. Across that whole interval $x - r_1$ holds one fixed sign and $x - r_2$ holds one fixed sign, because neither factor reaches its root there. A product of numbers with fixed signs has a fixed sign, since $(+)(+)$ and $(-)(-)$ are positive while $(+)(-)$ is negative. Multiplying by the constant $a$ afterwards only reverses or preserves that sign, uniformly across the entire interval. Therefore $f$ has one constant sign on the whole interval.

The only way $f$ can pass from positive to negative is for one of its factors to change sign, and a factor changes sign only as $x$ crosses its root. Between two consecutive roots no factor changes, so $f$ never changes there. The roots are the only places the sign can turn over, which is the claim.

The consequence is the entire method. Two roots cut the number line into three pieces, the stretch left of both, the stretch between them, and the stretch right of both. On each piece the sign of the quadratic is one fixed thing. So solving a quadratic inequality is never about checking infinitely many points. It is about finding the roots, which split the line, and then labelling each piece positive or negative.

Reading the sign: test a point, or read the parabola

Once you know the sign is constant on each piece, you need the sign on each one, and there are two ways to get it.

The first is a test point. Pick any convenient number inside a piece, substitute it, and record whether the result comes out positive or negative. That one value settles the entire piece, because the sign cannot change again until the next root. Testing $x = 0$ in $(x - 3)(x + 2)$ gives $(-3)(2) = -6 < 0$ , so the quadratic is negative across the whole middle piece.

The second way is faster and harder to get wrong: read the parabola. You already know that the sign of $a$ fixes which way a parabola opens. When $a > 0$ it opens upward, so it dips below the $x$ -axis between its two roots and rises above the axis outside them. So with no substituting at all, an upward parabola is negative exactly on the open interval between the roots and positive outside them. When $a < 0$ it opens downward and the picture flips top to bottom, positive between the roots and negative outside. The graph route is quicker because it computes nothing: you just recall the shape and the two roots. The graph route is also safer, because there is no arithmetic with signed numbers to slip on. Keep the test point as the backup for when you are unsure.

This is the same fact the discriminant already told you. For $a > 0$ with two real roots, the vertex sits at height $-\dfrac{\Delta}{4a}$ , which is negative, so the bottom of the parabola lies below the axis. That is precisely why an upward parabola with two roots is negative between them.

Whichever route you use, write the answer as a solution set, not as a loose phrase. The systems chapter made the point that the solution of an inequality is a set, an object you can name, draw, and combine. That solution is not merely a lone number to solve for. Here that set is a piece of the number line, written in interval notation, and when it arrives in two pieces you join them with a union $\cup$ . Shading it on a number line is just the one-dimensional version of shading a region in the plane.

Worked example 1 Solve $x^2 - x - 6 \ge 0$

The quadratic is already compared to zero, so factor it. Two numbers with product $-6$ and sum $-1$ are $-3$ and $2$ , giving

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

with roots $x = 3$ and $x = -2$ . These cut the line into three pieces.

The leading coefficient is $a = 1 > 0$ , so the parabola opens upward: negative between the roots, positive outside them. We want where it is greater than or equal to zero, which is the two outside pieces together with the roots themselves. Those roots are included, since the inequality is non-strict and they make the quadratic exactly zero:

x \le -2 \quad \text{or} \quad x \ge 3, \qquad \text{that is} \qquad (-\infty, -2] \cup [3, \infty).

The union is not decoration. The solution is two separate rays, not a single interval, and writing it as one interval would be wrong.

The parabola y = x squared minus x minus 6 opens upward and crosses the axis at the roots negative 2 and 3. It is at or above the axis outside the roots, so the solution of the inequality is the two shaded rays on the x-axis, with the roots filled in because the symbol allows equality.

Check your understanding

Solve $(x + 1)(x - 4) < 0$ .

(-1, 4)

[-1, 4]

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

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

Strict versus non-strict endpoints

The steps above decide which pieces of the line make up the answer. The inequality symbol decides only one more thing, what happens at the endpoints, which are the roots. This is trivial to state and constantly gotten wrong, so pin it down.

The roots are exactly the inputs where the quadratic equals zero. A strict symbol, $>$ or $<$ , asks for values that are honestly positive or honestly negative, so it excludes the roots. That means open circles on the number line, and parentheses in interval notation. A non-strict symbol, $\ge$ or $\le$ , also accepts zero, so it includes the roots: closed circles, square brackets. The interior of the solution is identical either way. Only the endpoints change. So $x^2 - x - 6 > 0$ has solution $(-\infty, -2) \cup (3, \infty)$ , the same two rays as the worked example but with the roots now removed.

When the leading coefficient is negative

A negative $a$ flips the parabola upside down, and with it every sign in the argument. There are two safe ways to handle it, and one tempting way that goes wrong.

The safe habit is to make $a$ positive first. Multiply both sides of the inequality by $-1$ , and remember the rule from the linear inequalities lesson: multiplying an inequality by a negative number reverses its direction. Now you are back to an upward parabola and the reading you already trust. The other safe route is to read the downward parabola directly. With $a < 0$ the parabola opens down, so it is positive between its roots and negative outside them, the mirror image of the upward case.

The tempting error is to multiply or divide by the negative and leave the symbol pointing the same way. That silently swaps the solution set for its opposite. If you ever rewrite an inequality by a negative factor, the symbol must turn around.

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

Here $a = -1$ , so use both routes and watch them agree.

Making $a$ positive, multiply through by $-1$ and reverse the symbol:

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

Factor the upward quadratic: $x^2 - 2x - 3 = (x - 3)(x + 1)$ , with roots $3$ and $-1$ . An upward parabola is at or below zero on the closed interval between its roots, so

-1 \le x \le 3, \qquad \text{that is} \qquad [-1, 3].

Reading the downward parabola directly gives the same thing. The roots of $-x^2 + 2x + 3$ are again $-1$ and $3$ , and because $a < 0$ the parabola opens downward and is positive between its roots. Including the roots for the non-strict symbol, the solution set is once more $[-1, 3]$ . The two routes must agree, because they describe the same object.

With a negative leading coefficient the parabola y = negative x squared plus 2x plus 3 opens downward and sits at or above the axis between its roots negative 1 and 3. The solution set is the single shaded segment between the roots, endpoints filled because the symbol allows equality.

Check your understanding

Solve $-x^2 - x + 6 > 0$ .

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

(-3, 2)

[-3, 2]

(-2, 3)

What the discriminant decides

The factored form $a(x - r_1)(x - r_2)$ assumed two real roots. Whether a quadratic has them is settled by the discriminant $\Delta = b^2 - 4ac$ from the quadratic formula. $\Delta > 0$ gives two real roots, $\Delta = 0$ gives one repeated root, and $\Delta < 0$ gives no real roots at all. Each case answers a quadratic inequality differently, and the two extreme cases are where both students and authors slip. Take $a > 0$ throughout, since you can always arrange that first.

Three upward parabolas, one for each sign of the discriminant. On the left, two real roots and a dip below the axis (negative between the roots, positive outside). In the middle, a repeated root where the curve only touches the axis (positive everywhere else). On the right, no real roots and the whole curve floating above the axis (positive for every input).

When $\Delta < 0$ there are no real roots, so the parabola never touches the axis. By the structural fact it can change sign nowhere, and since an upward parabola climbs to infinity it must be positive everywhere. There is a cleaner way to see it than “it has to be,” using the completed square.

Why a quadratic with no real roots keeps one sign#

Completing the square, the lesson before last, writes any quadratic in vertex form $f(x) = a(x - h)^2 + k$ , where the vertex height is $k = -\dfrac{\Delta}{4a}$ . Suppose $a > 0$ and $\Delta < 0$ . Then $k = -\dfrac{\Delta}{4a}$ is a negative divided by a positive with an overall minus sign, so $k > 0$ .

Now read off the sign. The square $(x - h)^2$ is never negative, and $a > 0$ , so $a(x - h)^2 \ge 0$ for every $x$ . Adding the positive number $k$ ,

f(x) = a(x - h)^2 + k \ge k > 0 \quad\text{for every } x.

So $f$ is strictly positive everywhere. There is no arithmetic with roots and no sign chart, only the fact that a real square cannot be negative. It follows at once that $f(x) > 0$ has solution set all of $\mathbb{R}$ , while $f(x) < 0$ has solution set $\varnothing$ , the empty set, because $f$ is never negative. With $a < 0$ the same argument gives $k < 0$ and $f(x) < 0$ everywhere, so the two answers swap.

Worked example 3 Solve $x^2 + x + 1 > 0$ , then $x^2 + x + 1 < 0$

Check the discriminant first: $\Delta = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0$ , so there are no real roots and no factoring over the real numbers. Complete the square instead:

x^2 + x + 1 = \left(x + \tfrac{1}{2}\right)^2 + \tfrac{3}{4}.

The square is never negative, so the whole expression is at least $\tfrac{3}{4}$ , which is positive, for every real $x$ . Therefore $x^2 + x + 1 > 0$ is true for all inputs, and its solution set is all of $\mathbb{R}$ . The opposite inequality $x^2 + x + 1 < 0$ is never true, so its solution set is the empty set $\varnothing$ . A negative discriminant is not a dead end. It is the case where the quadratic keeps one sign forever, and reading that sign answers the inequality immediately.

When $\Delta = 0$ the quadratic has one repeated root $r$ , and factoring gives a perfect square, $f(x) = a(x - r)^2$ . With $a > 0$ this is zero at $r$ and strictly positive everywhere else, since a nonzero square is positive. Notice this quadratic never actually changes sign: it only touches zero at $r$ and stays positive on both sides. The single quadratic then answers four inequalities four different ways, which makes it the sharpest test of whether you have the endpoints right.

Worked example 4 One repeated root, four answers

The quadratic $x^2 - 6x + 9$ has $\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$ , a repeated root. It is the perfect square

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

which equals zero only at $x = 3$ and is positive for every other $x$ . Run all four symbols against it:

(x - 3)^2 > 0 \;\Rightarrow\; (-\infty, 3) \cup (3, \infty), \qquad (x - 3)^2 \ge 0 \;\Rightarrow\; \mathbb{R},

(x - 3)^2 < 0 \;\Rightarrow\; \varnothing, \qquad (x - 3)^2 \le 0 \;\Rightarrow\; \{3\}.

Read them one at a time. The square is positive everywhere except at $3$ , so $> 0$ is the whole line with the single point $3$ punctured out, a union of two open rays. Allowing equality, $\ge 0$ picks up that point too and becomes all of $\mathbb{R}$ . Nothing makes a square negative, so $< 0$ is empty. And $\le 0$ holds only where the square is exactly zero, the single point $\{3\}$ . Four genuinely different solution sets from one quadratic, separated entirely by the symbol and the one repeated root.

Check your understanding

Solve $(x - 2)^2 > 0$ .

\varnothing

\mathbb{R}

\{2\}

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

Common mistakes

Multiplying by a negative and keeping the symbol. Rewriting $-x^2 + 2x + 3 \ge 0$ as $x^2 - 2x - 3 \ge 0$ leaves the symbol pointing the wrong way and hands back the opposite solution set. When you multiply or divide an inequality by a negative, the symbol must reverse. Safer still, make $a$ positive as the very first step.
Confusing strict and non-strict endpoints. $>$ and $<$ exclude the roots (open circles, parentheses); $\ge$ and $\le$ include them (closed circles, brackets). The intervals are the same either way, only the endpoints move.
Forcing the answer into a single interval. For an upward parabola with two roots, the “outside” inequality gives a union of two rays, such as $(-\infty, -2] \cup [3, \infty)$ , not one interval. Collapsing it to a single interval like $[-2, 3]$ reports the exact opposite set.
Believing “empty means no real roots.” Whether the solution is empty depends on the sign of $a$ , on $\Delta$ , and on the symbol, all three. For $a > 0$ and $\Delta < 0$ the inequality $f(x) < 0$ is empty, but $f(x) > 0$ is all of $\mathbb{R}$ . And $(x - 2)^2 < 0$ is empty even though the quadratic does have a real root. Never state the empty case with the hypotheses left out.
Treating a double root like a sign change. At a repeated root the quadratic touches zero without changing sign, so $(x - 3)^2 > 0$ removes only the single point $3$ and keeps both sides; it does not make an interval negative.
Testing a point that is actually a root. A test point sitting on a root gives zero, which decides nothing. Choose a point strictly inside a piece, never an endpoint.

Key takeaways

The structural fact. In factored form $a(x - r_1)(x - r_2)$ , each linear factor changes sign only at its own root, so their product changes sign only at $r_1$ or $r_2$ . Between consecutive roots the sign of the quadratic is constant.
Method. Move everything to one side, find the roots, and let them split the number line. Read the sign on each piece off the parabola, up for $a > 0$ and down for $a < 0$ , or confirm with one interior test point.
Answer as a set. Record the solution in interval notation, joining separate pieces with a union $\cup$ , exactly the solution-set object from the systems chapter, now living on the number line.
Endpoints. Strict $>$ , $<$ exclude the roots; non-strict $\ge$ , $\le$ include them. The intervals do not change, only whether the endpoints are open or closed.
The discriminant decides the shape of the answer. For $a > 0$ : if $\Delta > 0$ , the “inside” inequality gives one interval and the “outside” one gives a union of two rays. If $\Delta = 0$ , the single repeated root gives four different answers across the four symbols. And if $\Delta < 0$ , still with $a > 0$ , the quadratic is positive everywhere, so $> 0$ is all of $\mathbb{R}$ and $< 0$ is $\varnothing$ .

Quadratic Inequalities

Learning goals

A quadratic changes sign only at a root

Reading the sign: test a point, or read the parabola

Strict versus non-strict endpoints

When the leading coefficient is negative

What the discriminant decides

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)