Quadratics: Question 4

Worked solution

Part (a): Solving the quadratic inequality

First find the critical values by solving $2x^2 - 5x - 12 = 0$.

$\text{discriminant} = (-5)^2 - 4(2)(-12) = 25 + 96 = 121$

$x = \frac{5 \pm \sqrt{121}}{4} = \frac{5 \pm 11}{4}$

$x = \frac{16}{4} = 4 \qquad \text{or} \qquad x = \frac{-6}{4} = -\frac{3}{2}$

Since $a = 2 > 0$, the graph of $y = 2x^2-5x-12$ is an upward-opening parabola. It is below the $x$-axis between the roots and at or above the $x$-axis outside them. We want $2x^2-5x-12 \ge 0$, so:

$x \le -\frac{3}{2} \quad \text{or} \quad x \ge 4$

Part (b): Combining with a linear inequality

Solve the linear inequality on its own:

$3x - 1 < 11$

$3x < 12$

$x < 4$

Now find the values of $x$ satisfying both conditions. The intersection of $\left(x \le -\tfrac{3}{2} \text{ or } x \ge 4\right)$ with $\left(x < 4\right)$:

• The branch $x \le -\tfrac{3}{2}$ lies entirely inside $x<4$ (since $-\tfrac{3}{2} < 4$), so it survives unchanged.
• The branch $x \ge 4$ has no overlap with $x<4$ (the point $x=4$ itself is excluded, since the inequality is strict), so it is discarded entirely.

$x \le -\frac{3}{2}$

Final answers

• (a) $x \le \boxed{-\tfrac{3}{2}}$ or $x \ge \boxed{4}$
• (b) $x \le \boxed{-\tfrac{3}{2}}$