A quadratic inequality has the form \( ax^2 + bx + c \gt 0 \) (or with \( \lt , \geq, \leq \)), where \( a \neq 0 \). The function \( y = ax^2 + bx + c \) is a parabola.

The direction of opening depends on the leading coefficient \( a \):

• If \( a \gt 0 \) — the parabola opens upward ("smiling"). It is negative between the roots and positive outside them.
• If \( a \lt 0 \) — the parabola opens downward ("frowning"). It is positive between the roots and negative outside them.

The roots are the points where the expression equals zero, i.e., the solutions of \( ax^2 + bx + c = 0 \). Find them by factoring or by the quadratic formula:

Label the roots \( x_1 \lt x_2 \). They divide the number line into three intervals, and the sign of the expression is constant in each interval. It is therefore enough to test one point per interval, or simply use the shape of the parabola.

Note on endpoints: with a strict sign (\( \gt , \lt \)) the roots are not included in the solution (open dot ○). With a non-strict sign (\( \geq, \leq \)) the roots are included (filled dot ●), because the expression equals zero there.

Example 1: Upward Parabola — Solution Between the Roots

Problem: Solve the inequality: \( x^2 - 5x + 6 \leq 0 \)

Solution:

1. Find the roots: \( x^2 - 5x + 6 = 0 \). Factor: \( (x - 2)(x - 3) = 0 \).
2. The roots are \( x_1 = 2 \) and \( x_2 = 3 \).
3. The leading coefficient \( a = 1 \gt 0 \), so the parabola opens upward and is negative between the roots.
4. We need where the expression is less than or equal to zero (\( \leq 0 \)), i.e., between the roots including the endpoints.
5. Check: \( x = 2.5 \) gives \( 6.25 - 12.5 + 6 = -0.25 \lt 0 \). Indeed negative.

Answer: \( 2 \leq x \leq 3 \)

Example 2: Upward Parabola — Solution Outside the Roots

Problem: Solve the inequality: \( x^2 + 2x - 15 \gt 0 \)

Solution:

1. Find the roots: \( x^2 + 2x - 15 = 0 \). Factor: \( (x + 5)(x - 3) = 0 \).
2. The roots are \( x_1 = -5 \) and \( x_2 = 3 \).
3. The leading coefficient \( a = 1 \gt 0 \), so the parabola is positive outside the roots.
4. We need where the expression is strictly greater than zero (\( \gt 0 \)), i.e., outside the interval, not including the roots.
5. Check: \( x = 4 \) gives \( 16 + 8 - 15 = 9 \gt 0 \). Indeed positive.

Answer: \( x \lt -5 \) or \( x \gt 3 \)

Example 3: Negative Leading Coefficient — Reversed Picture

Problem: Solve the inequality: \( -x^2 + 4x + 5 \geq 0 \)

Solution:

1. For convenience, multiply both sides by \( -1 \) and flip the sign: \( x^2 - 4x - 5 \leq 0 \).
2. Find the roots: \( (x - 5)(x + 1) = 0 \), giving \( x_1 = -1 \) and \( x_2 = 5 \).
3. The leading coefficient is now positive, so the parabola opens upward and is negative between the roots.
4. We need \( \leq 0 \), so the solution is between the roots including the endpoints.
5. Check in the original: \( x = 0 \) gives \( -0 + 0 + 5 = 5 \geq 0 \). Indeed in the interval.

Answer: \( -1 \leq x \leq 5 \)

Example 4: Already-Factored Inequality

Problem: Solve the inequality: \( x(x - 4) \lt 0 \)

Solution:

1. The expression is already factored; the roots are read directly: \( x_1 = 0 \) and \( x_2 = 4 \).
2. Expanding gives \( x^2 - 4x \), so the leading coefficient \( a = 1 \gt 0 \) — parabola opens upward and is negative between the roots.
3. We need where the expression is strictly less than zero (\( \lt 0 \)), i.e., between the roots, not including them.
4. Check: \( x = 2 \) gives \( 2 \cdot (-2) = -4 \lt 0 \). Indeed negative.

Answer: \( 0 \lt x \lt 4 \)

Example 5: Inequality with No Linear Term

Problem: Solve the inequality: \( 2x^2 - 18 \geq 0 \)

Solution:

1. Factor out the common factor: \( 2(x^2 - 9) \geq 0 \), then divide by \( 2 \) (positive): \( x^2 - 9 \geq 0 \).
2. Factor as a difference of squares: \( (x - 3)(x + 3) \geq 0 \), so the roots are \( x_1 = -3 \) and \( x_2 = 3 \).
3. The leading coefficient is positive, so the parabola is positive outside the roots.
4. We need \( \geq 0 \), so the solution is outside the roots including the endpoints.
5. Check: \( x = 4 \) gives \( 2 \cdot 16 - 18 = 14 \geq 0 \). Indeed in the interval.

Answer: \( x \leq -3 \) or \( x \geq 3 \)

✗ Common mistake: Treating the inequality like an equation and stopping at the roots, giving an answer such as \( x = 2, x = 3 \) instead of an interval.

✓ The correct way: The roots are only the boundaries. The answer is an entire interval: determine where the expression is positive or negative using the parabola's direction, and write the appropriate interval.

✗ Common mistake: Ignoring a negative leading coefficient and treating the parabola as if it opens upward.

✓ The correct way: When the leading coefficient is negative the parabola opens downward, so it is positive between the roots and negative outside them — the opposite of the usual case. It helps to multiply by \( -1 \) and flip the sign to work with a positive leading coefficient.

✗ Common mistake: Including the roots in the solution for a strict sign (\( \gt , \lt \)), or excluding them for a non-strict sign (\( \geq, \leq \)).

✓ The correct way: The roots make the expression equal to zero. They are included only when the sign is \( \geq \) or \( \leq \) (filled dot), and excluded when the sign is \( \gt \) or \( \lt \) (open dot).

Key points for quadratic inequalities:

• Rearrange to zero, find the roots of \( ax^2 + bx + c = 0 \).
• Upward parabola (\( a \gt 0 \)): negative between the roots, positive outside them.
• Downward parabola (\( a \lt 0 \)): positive between the roots, negative outside them.
• Endpoints included only for \( \geq \) and \( \leq \) (filled dot); excluded for \( \gt \) and \( \lt \) (open dot).
• Quadratic formula: \( x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).