Find the set of values of \(x\) for which \[2x^2 - 5x - 3 \gt 0.\] (3) Find the set of values of \(x\) for which \[\frac{x + 4}{x - 2} \le 3.\] (5)

Two quadratic inequalities, 8 marks in total. The method for both is the same: get everything on one side, factorise, find the critical values, and then decide the sign of the expression in each region. A sketch of the parabola or a sign table is what makes the last step reliable. The second part carries the extra trap of a variable denominator.

(a) Solve \(2x^2 - 5x - 3 \gt 0\) [3 marks]

Factorise first: \[2x^2 - 5x - 3 = (2x + 1)(x - 3) \quad \textbf{M1}\] The M1 is for an attempt to factorise or to solve the quadratic; the formula or completing the square would earn it equally. The critical values are where the expression is zero: \[x = -\tfrac{1}{2} \quad \text{and} \quad x = 3 \quad \textbf{A1}\] Now the sign. The coefficient of \(x^2\) is positive, so the parabola opens upwards and lies above the axis outside its two roots and below between them. Since we want the expression to be positive, \[x \lt -\tfrac{1}{2} \quad \text{or} \quad x \gt 3 \quad \textbf{A1}\] The final A1 needs both regions and the word "or". Writing \(-\tfrac{1}{2} \gt x \gt 3\) is meaningless, because no number is simultaneously less than \(-\tfrac{1}{2}\) and greater than 3.

(b) Solve \(\dfrac{x + 4}{x - 2} \le 3\) [5 marks]

The one thing you must not do is multiply both sides by \((x - 2)\), because that expression changes sign at \(x = 2\) and multiplying an inequality by a negative quantity reverses it. The safe manoeuvre is to multiply by \((x - 2)^2\), which is positive for every \(x \ne 2\) and so leaves the inequality sign alone. Record the excluded value first: \(x \ne 2\), since the original expression is undefined there.

\[(x + 4)(x - 2) \le 3(x - 2)^2 \quad \textbf{M1}\] \[x^2 + 2x - 8 \le 3x^2 - 12x + 12 \quad \textbf{A1}\] \[0 \le 2x^2 - 14x + 20 \ \Rightarrow\ x^2 - 7x + 10 \ge 0 \quad \textbf{A1}\] \[(x - 2)(x - 5) \ge 0 \ \Rightarrow\ x \le 2 \ \text{or} \ x \ge 5 \quad \textbf{M1}\]

Finally impose the exclusion. The value \(x = 2\) satisfies \((x-2)(x-5) \ge 0\) but is not in the domain of the original inequality, so it must be removed: \[x \lt 2 \quad \text{or} \quad x \ge 5 \quad \textbf{A1}\] The second M1 is for solving the new quadratic inequality correctly in structure; the last A1 is reserved for the fully correct final set, including the strict inequality at \(x = 2\) and the inclusive one at \(x = 5\). The A marks are not follow-through, so an expansion slip earlier costs them.

An alternative route that also earns full marks is to subtract 3 and combine into a single fraction: \[\frac{x+4}{x-2} - 3 = \frac{x + 4 - 3(x-2)}{x-2} = \frac{10 - 2x}{x - 2} \le 0,\] then use a sign table on the critical values \(x = 2\) and \(x = 5\). Both routes must end with \(x = 2\) excluded.

Common wrong turns on this question

• Multiplying by \((x - 2)\) in part (b) and obtaining only \(x \ge 5\). This is the error the question is built to catch, and it silently discards the whole region to the left of 2.
• Forgetting to exclude \(x = 2\), giving \(x \le 2\) and so claiming a solution where the expression is undefined.
• Cancelling the factor \((x - 2)\) from \((x+4)(x-2) \le 3(x-2)^2\). Dividing by a quantity that may be negative or zero is not permitted.
• Getting the direction wrong in part (a) by assuming "greater than zero" means "between the roots". Sketch the parabola, or test one convenient value.

How to check

Test one value from each region. In part (a), \(x = 0\) gives \(-3\), which is not positive, correctly excluding the middle region; \(x = 4\) gives \(9 \gt 0\). In part (b), \(x = 0\) gives \(\dfrac{4}{-2} = -2 \le 3\), so \(x \lt 2\) belongs; \(x = 3\) gives \(\dfrac{7}{1} = 7\), which is not at most 3, so the middle region is correctly excluded; \(x = 5\) gives exactly \(\dfrac{9}{3} = 3\), so the boundary is correctly inclusive. Three substitutions settle a whole inequality.