Solving a Quadratic Inequality

The reliable way to solve a quadratic inequality is not a sign table, it is a quick sketch. Factorise to find where the parabola cuts the axis, decide which way it opens, then read off the region that satisfies the inequality. This avoids the single most common error: giving “between” when the answer is “outside”.

Solve the inequality $2x^2 - 5x - 3 > 0$. [4]

The working

Step 1, factorise the quadratic (find the roots of $2x^2 - 5x - 3 = 0$): $2x^2 - 5x - 3 = (2x + 1)(x - 3) = 0 \;\Rightarrow\; x = -\tfrac{1}{2} \ \text{ and } \ x = 3 \quad \text{(M1, A1)}$

Step 2, sketch. The coefficient of $x^2$ is $+2 > 0$, so the parabola opens upward and cuts the $x$-axis at $-\frac12$ and $3$. Between the roots it dips below the axis; outside them it is above.

Step 3, read the region. We want $> 0$, i.e. where the curve is above the axis, the two outer pieces: $x < -\frac{1}{2} \quad \text{or} \quad x > 3 \quad \text{(M1, A1)}$

Where the marks are won and lost

• The direction of opening decides everything. With a positive $x^2$ coefficient and "$> 0$", the solution is the outside pair of regions. Getting this backwards ("$-\frac12 < x < 3$") is the classic lost mark.
• Use “or” between the two regions, not “and”. No single $x$ can be both less than $-\frac12$ and greater than $3$; writing it as a chained inequality $3 < x < -\frac12$ is meaningless.
• Strict "$>$" gives open regions (roots excluded). If it were "$\ge$", the roots would be included with $\le$ and $\ge$.

Common mistakes

• Giving the region between the roots when the answer is outside (or vice versa).
• Writing the answer as a single impossible inequality instead of two separate regions joined by “or”.
• Factorising sign-slips: $(2x + 1)(x - 3)$ expands to $2x^2 - 5x - 3$, check by expanding before trusting the roots.

Full method: Quadratic Inequalities notes. Topic home: Quadratic Functions pillar.