Quadratic Inequalities: Sign Charts and Endpoint Rules

You factorise a quadratic correctly, find both roots, and still lose the answer. Usually, the trouble starts just after the algebra: should you choose the space between the roots or the two outside stretches? And do the roots themselves belong?

A quadratic inequality asks for a set of values, not just the values that make an expression zero. The method of intervals helps you find that set without guessing.

We will work through ordinary factors, a negative leading coefficient, awkward roots, repeated roots and a small profit problem. Along the way, we will keep two decisions separate: which intervals work, and which boundary points work.

What are we trying to find?

Compare these two questions:

• Solve x² - 3x - 10 = 0.
• Solve x² - 3x - 10 < 0.

The equation asks where the expression is exactly zero. Its answers are x = -2 and x = 5.

The inequality asks where the expression is negative. Try x = 0: the expression becomes -10, so 0 works even though it is not a root. Try x = 6: the expression becomes 8, so 6 does not work.

The full inequality solution is -2 < x < 5. Every real number in this open interval works, including fractions and decimals. The endpoints do not, because zero is not less than zero.

We assume x is real unless a question gives a different domain. A requirement such as “x is an integer” or “x represents the number of items” must be applied to the final interval.

The method of intervals in five steps

1. Put zero on one side. Combine all terms into one quadratic expression.
2. Find its real zeros. Factorise, complete the square or use the quadratic formula.
3. Mark the distinct zeros in increasing order. They split the number line into open intervals.
4. Find the sign on each interval. Use a convenient test value or the signs of the factors.
5. Select intervals and check endpoints. Write the resulting set clearly.

This is the algebraic sign-chart approach described in OpenStax’s quadratic inequalities lesson. The examples below use our own numbers and checks.

Why can one test value represent a whole interval? A polynomial is continuous. To get from a positive value to a negative value, its graph must pass through zero. Once all real zeros have been marked, there is no unmarked zero inside any of the intervals where a sign change could happen.

Example 1: Build a complete sign chart

Solve x² - 3x - 10 < 0.

Factorise:

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

The zeros are -2 and 5. Use -3, 0 and 6 as test values for the three intervals.

Interval	x + 2	x - 5	Product
x < -2	Negative	Negative	Positive
-2 < x < 5	Positive	Negative	Negative
x > 5	Positive	Positive	Positive

We need a negative product, so choose the middle interval:

Answer: -2 < x < 5, or (-2, 5).

At x = -2 and x = 5, the product equals zero. Both endpoints are excluded because the inequality is strict.

Notice something useful: both factors are negative on the left, yet their product is positive. Looking at only one factor would give the wrong answer.

Change only the inequality symbol

Keep the same expression and change the condition:

Condition	Solution
x² - 3x - 10 < 0	(-2, 5)
x² - 3x - 10 ≤ 0	[-2, 5]
x² - 3x - 10 > 0	(-∞, -2) ∪ (5, ∞)
x² - 3x - 10 ≥ 0	(-∞, -2] ∪ [5, ∞)

The signs on the intervals have not changed. Only the intervals we select and the treatment of the endpoints change.

Read interval notation without losing marks

A round bracket means the finite endpoint is excluded. A square bracket means it is included.

• (2, 7) means 2 < x < 7.
• [2, 7] means 2 ≤ x ≤ 7.
• [2, 7) means 2 ≤ x < 7.
• (-∞, 2] ∪ [7, ∞) means x ≤ 2 or x ≥ 7.

The symbol ∪ joins alternative sets. For two outside regions, use or, not and. No real number can be both less than 2 and greater than 7 at the same time.

Infinity always takes a round bracket. It is not a real endpoint that you can include.

On a number line, use a hollow circle for an excluded endpoint and a filled circle for an included one. Shade only the selected regions. For a singleton answer such as {4}, mark just that one point.

Example 2: Move everything before finding the roots

Solve 2x² + x ≤ 15.

Subtract 15 from both sides:

2x² + x - 15 ≤ 0

Now factorise:

(2x - 5)(x + 3) ≤ 0

The zeros are 5/2 and -3. Put them in numerical order: -3, 5/2.

On the left of -3, both factors are negative, giving a positive product. Between -3 and 5/2, the factors have opposite signs, giving a negative product. To the right of 5/2, both are positive.

Choose the middle interval and include both zeros:

Answer: [-3, 5/2].

Check an interior value in the original question: x = 0 gives 0 ≤ 15, which is true. Check the upper endpoint: x = 5/2 gives 2(25/4) + 5/2 = 15, so equality really does hold.

If you had solved 2x² + x = 0 instead, you would have found the wrong boundaries. The boundaries belong to the expression after both sides have been combined.

Example 3: Handle a negative leading coefficient

Solve -2x² + 7x + 4 ≥ 0.

There are two valid routes. You can retain the negative leading coefficient in your sign chart, or multiply the whole inequality by -1.

Using the second route:

2x² - 7x - 4 ≤ 0

The inequality reverses because we multiplied both sides by a negative number.

Factorise:

(2x + 1)(x - 4) ≤ 0

The roots are -1/2 and 4. The new quadratic is negative between its roots and zero at them.

Answer: [-1/2, 4].

Check x = 0 in the original expression: 4 ≥ 0 is true. Check x = 5: -50 + 35 + 4 = -11, which fails.

If you keep the original expression, its sign pattern is negative, positive, negative. That produces exactly the same answer.

Why the inside-or-outside shortcut works

Suppose a quadratic has two distinct real roots r and s, with r < s. Write it as a(x - r)(x - s).

Outside the roots, the two brackets have matching signs, so their product is positive. Between the roots, they have opposite signs, so their product is negative. The coefficient a then either preserves those signs or reverses them.

A quick graph tells the same story. With a > 0, the parabola opens upward and lies below the horizontal axis between its two crossings. With a < 0, it opens downward and lies above the axis between those crossings.

This shortcut requires two distinct real roots. Do not apply it mechanically when the discriminant is zero or negative.

Example 4: Roots do not have to be neat integers

Solve x² - 4x + 1 > 0.

Completing the square gives:

(x - 2)² - 3 > 0

The zeros are 2 - √3 and 2 + √3. Because the leading coefficient is positive, the expression is positive outside the two roots.

Answer: (-∞, 2 - √3) ∪ (2 + √3, ∞).

Both endpoints are excluded.

You can use approximate roots, about 0.268 and 3.732, to place them on a rough sketch. Keep the exact surds in your final answer. Rounding a boundary to 0.27 changes which nearby numbers the answer includes.

For instance, 0.269 is larger than the true lower root, so it lies in the negative region. A rounded condition x < 0.27 would wrongly include it.

Example 5: A repeated root does not flip the sign

Consider x² - 8x + 16 = (x - 4)².

There is only one distinct zero: 4. On either side, the square is positive. At 4, it is zero.

That single observation settles all four inequalities:

Condition	Solution
(x - 4)² > 0	(-∞, 4) ∪ (4, ∞)
(x - 4)² ≥ 0	All real numbers
(x - 4)² < 0	No real solution
(x - 4)² ≤ 0	{4}

Try x = 3 and x = 5: both give 1. The graph touches the axis and turns back instead of crossing it. This is the even-multiplicity behaviour explained in OpenStax’s discussion of polynomial zeros.

A common mistake is to see a root and automatically alternate signs. At a double root, both identical factors change sign together, leaving their product positive on both sides.

For -(x - 4)², the expression is negative everywhere except at 4. Thus -(x - 4)² ≥ 0 has the answer {4}, while -(x - 4)² < 0 excludes only 4.

Example 6: No real roots can mean every value works

Solve x² + 6x + 13 > 0.

Complete the square:

x² + 6x + 13 = (x + 3)² + 4

A real square is never negative. Adding 4 makes this expression strictly positive for every real x.

Answer: all real numbers, written (-∞, ∞).

For the same expression, the inequality x² + 6x + 13 ≤ 0 has no real solution.

“No real roots” tells us there are no places where the graph meets the axis. It does not tell us that an inequality has no answer.

More generally, for ax² + bx + c with a ≠ 0, calculate D = b² - 4ac:

• D > 0: two distinct real roots; use three intervals.
• D = 0: one distinct repeated root; check the square and the zero separately.
• D < 0: no real roots; the expression keeps the sign of a everywhere.

The discriminant cases and interval method are also developed in this algebra teaching module. Treat the leading coefficient as part of the sign analysis every time.

Example 7: Add a practical domain restriction

A school fair stall models its profit by P(q) = -q² + 18q - 65, where P is in hundreds of rupees and q is a whole number of batches sold. Assume the model is used only for 0 ≤ q ≤ 18.

How many batches must be sold to make a strictly positive profit?

Solve:

-q² + 18q - 65 > 0

Multiply by -1 and reverse the sign:

q² - 18q + 65 < 0

Factorise:

(q - 5)(q - 13) < 0

The real solution is 5 < q < 13. Since q must be a whole number, the permitted quantities are:

6, 7, 8, 9, 10, 11 or 12 batches.

Selling 5 or 13 batches gives zero profit, so neither meets the strict requirement. If the question asked for no loss, the condition would be P(q) ≥ 0 and the permitted quantities would run from 5 through 13 inclusive.

For a numerical check, at q = 9 the model gives P = -81 + 162 - 65 = 16, meaning Rs 1,600 profit. At q = 14 it gives -9, meaning a loss of Rs 900.

The algebra produces an interval. The meaning of the variable then narrows it to usable answers.

Example 8: Two conditions must overlap

Solve x² - 3x - 10 ≤ 0 and x > 1.

The quadratic condition gives [-2, 5]. The second condition gives (1, ∞).

Because both conditions must hold, take their intersection, the part they share:

Answer: (1, 5].

The left endpoint is open because x > 1 excludes 1. The right endpoint is closed because 5 satisfies both conditions.

Mixed brackets often appear when restrictions are combined. You do not need to force both ends to look alike.

Five mistakes worth catching before you submit

1. Stopping at the roots

Writing “x = -2, 5” answers an equation. For an inequality, continue to a sign chart and a solution set.

2. Dividing by an expression whose sign is unknown

From x(x - 4) > 0, dividing by x and writing x - 4 > 0 loses all solutions with x < 0. For negative x, division must reverse the sign. At x = 0, division is not allowed at all.

The correct answer is x < 0 or x > 4. Keeping the product intact avoids unnecessary cases.

3. Assuming a positive product needs two positive factors

Two negative factors also give a positive product. In (x + 2)(x - 5) > 0, the left outside interval is just as important as the right one.

4. Including roots in a strict inequality

A root gives zero. Zero satisfies ≥ 0 and ≤ 0, but it satisfies neither > 0 nor < 0. Substitute the endpoint if you are uncertain.

5. Reusing polynomial endpoint rules for fractions

In a polynomial inequality, real zeros are valid inputs. In a rational inequality, a denominator equal to zero makes the expression undefined. Such a value stays excluded even when the symbol includes equality.

A short practice set

Try these without looking at the answers. For each question, write the roots or explain why there are no real roots, then show the selected set.

1. x² + x - 12 < 0
2. 3x² - 8x - 3 ≥ 0
3. -x² + 2x + 15 > 0
4. (x + 2)² ≤ 0
5. (x + 2)² > 0
6. x² - 2x + 5 < 0
7. x² - 2x - 1 ≤ 0
8. x² - 7x + 10 < 0, where x is an integer
9. x² - 9 ≥ 0 and x < 4
10. -3(x - 1)² ≤ 0

Answers with the key reason

1. (-4, 3). Factorise as (x + 4)(x - 3). Choose the negative middle interval, excluding both roots.
2. (-∞, -1/3] ∪ [3, ∞). Factorise as (3x + 1)(x - 3). Choose the positive outside intervals and include zero-valued boundaries.
3. (-3, 5). The downward-opening quadratic has roots -3 and 5 and is positive between them.
4. {-2}. A square can be at most zero only when it equals zero.
5. (-∞, -2) ∪ (-2, ∞). The square is positive everywhere except at its zero.
6. No real solution. The expression is (x - 1)² + 4, which is always positive.
7. [1 - √2, 1 + √2]. The expression is non-positive between its two roots, including them.
8. {3, 4}. The real interval is (2, 5); retain only its integers.
9. (-∞, -3] ∪ [3, 4). Intersect the two outside regions with x < 4.
10. All real numbers. A square multiplied by -3 is always non-positive, including at x = 1.

If your answer differs, identify whether the error came from factorisation, interval signs or endpoints. Practising the exact weak step is more useful than repeating a whole page of questions blindly.

Frequently asked questions

What is the method of intervals for quadratic inequalities?

It means locating the real zeros, using them to divide the number line, checking the expression’s sign in each region, and choosing the regions that satisfy the inequality. Endpoints are checked separately.

Do I always choose the interval between the roots?

No. For a positive leading coefficient and two distinct real roots, the quadratic is negative between the roots and positive outside. A negative leading coefficient reverses this. Match the sign to the actual question.

When do I use square brackets?

Use them at a finite endpoint that belongs to the solution. A polynomial zero is included for ≤ 0 or ≥ 0, provided any other domain restrictions also allow it. Infinity never takes a square bracket.

Why does multiplying by -1 reverse an inequality?

Multiplication by a negative number reverses order on the number line. For example, 2 < 5 becomes -2 > -5. Apply the reversal to the whole inequality, not to individual terms separately.

Can I use the quadratic formula instead of factorisation?

Yes. Solve ax² + bx + c = 0 using x = (-b ± √(b² - 4ac))/(2a). If there are two real roots, put them in increasing order and use them as the interval boundaries.

Does the sign change at every root?

Not at a repeated quadratic root. In a(x - r)², the square has the same positive sign on both sides of r. Only its value at r is zero.

What if the discriminant is negative?

There are no real zeros, so a genuine quadratic keeps the sign of its leading coefficient for every real input. Depending on the requested inequality, the solution is all real numbers or no real numbers.

Can a quadratic inequality have just one answer?

Yes. For example, (x - 4)² ≤ 0 holds only at x = 4. Write {4}, not an empty open interval (4, 4).

Is checking x = 0 enough?

Only for the interval that contains 0, and only if 0 is not a root. With two distinct roots, check each interval or use a justified factor-sign pattern.

What should I write if there is no solution?

Write “no real solution” or the empty-set symbol ∅. Do not write x = 0 unless zero itself satisfies the inequality; a zero value and an empty solution set mean different things.