Quadratic Inequalities

Finding Critical Values

The critical values are the exact points where the quadratic expression equals zero. They mark the strict boundaries of your final solution set.

To find them, you temporarily replace the inequality sign with an equals sign to form the equation $ax^2 + bx + c = 0$. You then solve this quadratic equation, usually by factorising into the form $(x \pm p)(x \pm q) = 0$. If the quadratic cannot be factorised, you will need to use the quadratic formula or complete the square to find the critical values.

Visualising with a Parabola

Once you have your critical values, you must sketch a parabola. This sketch is a vital visual tool to determine which $x$-values satisfy the inequality.

• For $> 0$ or $\geq 0$: The solution lies in the regions above the $x$-axis. This forms the two outer "tails" of the parabola, creating disjoint intervals.
• For $< 0$ or $\leq 0$: The solution lies in the region the -axis. This forms the single "valley" between the critical values, creating one continuous interval.

Disjoint intervals do NOT overlap and cannot mathematically be written as a single combined string (e.g., $5 < x < -2$ is forbidden). They must be separated by the word "or".

Representing the Solution Set

The solution set can be represented algebraically, on a number line, or on a graph using formal conventions:

• Number Lines: Use open circles for strict inequalities ($<, >$) and solid closed circles for inclusive inequalities ($\leq, \geq$).
• Graphs: Use dashed lines for strict inequalities and solid lines for inclusive ones. AQA examiner guidance advises shading the non-required (unwanted) region so the actual solution region remains clear.
• Set Notation: This is a formal way of writing the solution set. It uses curly brackets { } and a colon : (meaning "such that"). Use the union symbol $\cup$ for "or" (disjoint regions) and the intersection symbol $\cap$ for "and".

Worked Example 1: Disjoint Regions ($\geq 0$)

Calculate the solution set for $x^2 - x - 12 \geq 0$, giving your answer in set notation.

Step 1: Set to zero to find critical values.

• $x^2 - x - 12 = 0$

Step 2: Factorise and solve.

• $(x - 4)(x + 3) = 0$
• The critical values are $x = 4$ and $x = -3$.

Step 3: Sketch the parabola and identify the region.

• Draw a positive (U-shaped) parabola crossing the $x$-axis at $-3$ and $4$.
• Because the inequality is $\geq 0$, look for the regions on or above the $x$-axis.
• These are the two outer "tails", which are disjoint.

Step 4: Express the final solution set.

• Inequality notation: $x \leq -3$ or $x \geq 4$.
• Set notation: $\{x : x \leq -3\} \cup \{x : x \geq 4\}$.

Worked Example 2: Single Interval ($< 0$)

Calculate the solution set for $x^2 - 3x - 10 < 0$.

Step 1: Set to zero to find critical values.

• $x^2 - 3x - 10 = 0$

Step 2: Factorise and solve.

• $(x - 5)(x + 2) = 0$
• The critical values are $x = 5$ and $x = -2$.

Step 3: Sketch the parabola and identify the region.

• Draw a positive parabola crossing the $x$-axis at $-2$ and $5$.
• Because the inequality is $< 0$, look for the region below the $x$-axis.
• This is the single "valley" between the two critical values.

Step 4: Express the final solution set.

• Inequality notation: $-2 < x < 5$.
• Set notation: $\{x : -2 < x < 5\}$.